Texas Go Math

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 5.2 Answer Key Experimental Probability of Simple Events.

Texas Go Math Grade 7 Lesson 5.2 Answer Key Experimental Probability of Simple Events

Texas Go Math Grade 7 Lesson 5.2 Explore Activity Answer Key

Finding Experimental Probability

You can toss a paper cup to demonstrate experimental probability.
A. Consider tossing a paper cup. Fill in the Outcome column of the table with the three different ways the cup could land.

B. Toss a paper cup twenty times. Record your observations in the table.

Reflect

Question 1.
Which outcome do you think is most likely?
Answer:
The most likely outcome is the one that happens most often.

Go Math Grade 7 Lesson 5.2 Answer Key Question 2.
Describe the three outcomes using the words likely and unlikely.
Answer:
Outcome – Open-ended up an experiment is unlikely;
Outcome – open-end down is unlikely;
Outcome – its side is likely.

Question 3.
Use the number of times each event occurred to calculate the probability of each event.
Answer:

P (open-end up) = \(\frac{4}{20}\)
P (open-end down) = \(\frac{4}{20}\)
P (on its side) = \(\frac{4}{20}\)

Question 4.
What do you think would happen if you performed more trials?

Answer:
Probability will slowly approximate.

Question 5.
What is the sum of the three probabilities in the table?
Answer:
The sum of the three probabilities is:
\(\frac{3}{20}\) + \(\frac{4}{20}\) + \(\frac{13}{20}\) = 1

Example 1

Martin has a bag of marbles. He removed one marble, recorded the color, and then placed it back in the bag. He repeated this process several times and recorded his results in the table. Find the probability of drawing each color. Write your answers in the simplest form.

A. Number of trials = 50
B. Complete the table of experimental probabilities. Write each answer as a fraction in simplest form.

Reflect

Lesson 5.2 Probability of Simple Events Answer Key Question 6.
What are two different ways you could find the experimental probability of the event that you do not draw a red marble?
Answer:
P(red) = \(\frac{6}{25}\)
P(blue) = \(\frac{1}{5}\)
P(green) = \(\frac{3}{10}\)
P(yellow) = \(\frac{13}{50}\)
First Way
The sum of the probabilities of an event and ¡ts complement equals 1
P(event) + P(complement) = 1
P(red) + P(not red) = 1 Substitute \(\frac{6}{25}\) for P(red).
\(\frac{6}{25}\) + P(not red) = 1 Subtract \(\frac{6}{25}\) from both sides.
P(not red) = 1 – \(\frac{6}{25}\)
P(not red) = \(\frac{19}{25}\)
The probability of not drawing a red marble is \(\frac{19}{25}\).

Second Way
The probability of not drawing a red marble equals the probability of drawing a marble of any other color.

Since we have 3 other colors, the probability of drawing a marble of any other color is the sum of the probabilities of drawing each one of them.

P(not red) = P(blue) + P(green) + P(yellow) Substitute \(\frac{1}{5}\) for P(blue), \(\frac{3}{10}\) for P(green), and \(\frac{13}{50}\).
P(not red) = \(\frac{1}{5}+\frac{3}{10}+\frac{13}{50}\)
P(not red) = \(\frac{1 \cdot 10+3 \cdot 5+13}{50}\)
P(not red) = \(\frac{38}{50}\)
P(not red) = \(\frac{19 \cdot 2}{25 \cdot 2}\)
P(not red) = \(\frac{19}{25}\)
The probability of not drawing a red marble is \(\frac{19}{25}\).

Your Turn

Question 7.
A spinner has three unequal sections: red, yellow, and blue, The table shows the results of Nolan’s spins. Find the experimental probability of landing on each color. Write your answers in the simplest form.

Answer:
The number of trials is the sum of frequencies.
Total number of trials is = 10 + 14 + 6 = 30
P(red color) = Substitute 10 for frequency of the event, and 30 for total number of trials.
P(red color) = \(\frac{10}{30}\)
P(red color) = \(\frac{1 \cdot 10}{3 \cdot 10}\)
P(red color) = \(\frac{1}{3}\)

P(yellow color) = Substitute 14 for frequency of the event, and 30 for total number of trials.
P(yellow color) = \(\frac{14}{30}\)
P(yellow color) = \(\frac{7 \cdot 2}{15 \cdot 2}\)
P(yellow color) = \(\frac{7}{15}\)

P(blue color) = Substitute 6 for frequency of the event, and 30 for total number of trials.
P(yellow color) = \(\frac{6}{30}\)
P(yellow color) = \(\frac{1 \cdot 6}{5 \cdot 6}\)
P(yellow color) = \(\frac{1}{5}\)

Experimental Probability Answer Key Go Math Lesson 5.2 Question 8.
A toy machine has equal numbers of red, white, and blue rubber balls. Ross wonders which color ball will come out of the machine next. Describe how you can use a standard number cube to model this situation. Then use a simulation to predict the color of the next ball.
Answer:
The standard cube has six sides marked with numbers 1, 2, 3, 4, 5, 6. If the numbers 1 and 2 come up, the red ball is released, 3 or 4 comes out white and 4 or 5 turns out blue.
Let’s throw the cube 10 times and we’ll make numbers:
6, 5, 5, 4, 5, 2, 3, 1, 2, 4
It means it will. come out 4 blue, 3 white, and 3 red balls.
Therefore, you can predict that the blue ball has the most of chance coming out.

Texas Go Math Grade 7 Lesson 5.2 Guided Practice Answer Key

Question 1.
Toss a coin at least 20 times. (Explore Activity and Example 1)

a. Record the results in the table.
Answer:

P (flip a head) = \(\frac{2}{5}\)
P (flip a tails) = \(\frac{3}{5}\)

b. What do you think would happen if you performed more trials?
Answer:
Probabilities will be the same.

Question 2.
Rachel’s free-throw average for basketball is 60%. Describe how you can use 10 index cards to model this situation. Then use a simulation to predict how many times in the next 50 tries Rachel will make a free throw. (Example 2).
Answer:
Let the cards be marked with numbers from 1 to 10. Let numbers 1, 2, 3, 4, 5, 6 represent that free-throw is on average, while numbers 7, 8, 9, lO, and the free-throw was beLow the average.

Let’s draw one card, we’ll record the number that we have and return it again Let’s repeat this 50 times
4, 7, 5, 6, 8, 5, 4, 2, 1, 1, 1, 9, 7, 10, 10, 6, 3, 4, 4, 5, 7, 8, 2, 1, 5, 8, 4, 10, 2, 2, 6, 7, 7, 5, 4, 1, 3, 2, 2, 2, 6, 7, 10, 3, 5, 9, 7, 4, 4, 1
Rachel will make a free throw 34 times.

Essential Question Check-In

Question 3.
Essential Question Follow Up How do you find the experimental probability of a simple event?
Answer:
Experimental probability is a value of number of time the event occurs divided by total number of trials.

Texas Go Math Grade 7 Lesson 5.2 Independent Practice Answer Key

Question 4.
Dree rolls a strike in 6 out of the 10 frames of bowling. What is the experimental probability that Dree will roll a strike in the first frame of the next game? Explain why a number cube would not be a good way to simulate this situation.
Answer:
P (roll a strike) =
The experimental probability that Dree will roll a strike in the first frame of the next game is \(\frac{3}{5}\).

The cube is not good for presenting this situation because we have the probability of achieving the goal of \(\frac{3}{5}\), so three numbers would be marked as winnings, two that were not, and with the remaining number would have no meaning.

Texas Go Math Grade 7 Answer Key Lesson 5.2 Probability Question 5.
To play a game, you spin a spinner like the one shown. You win if the arrow lands in one of the areas marked “WIN”. Lee played this game many times and recorded her results. She won 8 times and lost 40 times. Use Lee’s data to explain how to find the experimental probability of winning this game.

Answer:

P (lend on “WIN”) = \(\frac{1}{6}\)

Question 6.
Critique Reasoning A meteorologist reports an 80% chance of precipitation. Is this an example of experimental probability, written as a percent? Explain your reasoning.
Answer:
80% = \(\frac{80}{100}\) = \(\frac{4}{5}\)
This is the experimental probability because when we translate it into a fraction, it actually means that 4 out of 5 days will be precipitation.

Question 7.
The names of the students in Mr. Hayes’ math class are written on the board. Mr. Hayes writes each name on an index card and shuffles the cards. Each day he randomly draws a card, and the chosen student explains a math problem on the board. What is the probability that Ryan is chosen today? What is the probability that Ryan is not chosen today?

Answer:

P (Ryan is chosen) = \(\frac{1}{20}\)
P (Ryan is not chosen) = \(\frac{19}{20}\)

Probability Grade 7 Pdf Lesson 5.2 Answer Key Question 8.
Mica and Joan are on the same softball team. Mica got 8 hits out of 48 times at bat, while Joan got 12 hits out of 40 times at bat. Who do you think is more likely to get a hit her next time at bat? Explain.
Answer:

Is more likely that Joan get a hit because \(\frac{1}{6}\) < \(\frac{3}{10}\).

Question 9.
Make a Prediction In tennis, Gabby serves an ace, a ball that can’t be returned, 4 out of the 10 times she serves. What is the experimental probability that Gabby will serve an ace on the first serve of the next game? Make a prediction about how many aces Gabby will make on her next 40 serves. Justify your reasoning.
Answer:

The probability that Gabby will serve an ace is \(\frac{2}{5}\).
You can predict that she will serve an ace about 16 times out of 40.

H.O.T. Focus on Higher Order Thinking

Question 10.
Represent Real-World Problems Patricia finds that the experimental probability of her dog wanting to go outside between 4 p.m. and 5 p.m. is \(\frac{7}{12}\) About what percent of the time does her dog not want to go out between 4 p.m. and 5 p.m.?
Answer:
P (dog wanting to go outside) = \(\frac{7}{12}\)
P (dog not want to go outside) = 1 – P (dog wanting to go outside) = 1 – \(\frac{7}{12}\) = \(\frac{12}{12}\) – \(\frac{7}{12}\) = \(\frac{5}{12}\)
Use proportion.

About 41.7 % her dog not want to go out between 4 p.m. and 5 p.m.

Question 11.
Critique Reasoning Talia tossed a penny many times; she got 40 heads and 60 tails. She said the experimental probability of getting heads was \(\frac{40}{60}\). Explain the error and correct the experimental probability.
Answer:
She actually tossed the coin 100 times and out of 100 she got 40 heads and 60 tails.
P (getting heads) =
P (getting heads) = \(\frac{2}{5}\)

Lesson 5.2 Answer Key 7th Grade Experimental Probability Question 12.
Communicate Mathematical Ideas A high school has 438 students, with about the same number of males as females. Describe a simulation to predict how many of the first 50 students who leave school at the end of the day are female.
Answer:
P (student is a female) =
Have 25 red and 25 blue balls in the basket. If we choose a red ball, a female student comes out, otherwise a male student comes out.

Question 13.
Critical Thinking For a scavenger hunt, Chessa put one coin in each of 10 small boxes. Four coins are quarters, 4 are dimes, and 2 are nickels. How could you simulate choosing one box at random? What problem would there be if you planned to put these coins in your pocket and pick one?
Answer:
In the box are 10 balls, 4 blue, 4 white and 2 green. We choose randomly one ball If the ball is blue, we select one of the boxes in which the dimes are, if the ball is white, we select one of the boxes with quarter and if the ball is green we select one of the boxes with nickle.

If you put all the coins in your pocket then the probability will be equal for each coin.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 5 Quiz Answer Key.

Texas Go Math Grade 7 Module 5 Quiz Answer Key

Texas Go Math Grade 7 Module 5 Ready to Go On? Answer Key

5.1 Probability

Question 1.
Josue tosses a coin and spins the spinner at the right. What are all the possible outcomes? __________________

Answer:
The possible outcomes are;
1. head and section 1 (red);
2. head and section 2 (green);
3. tails and section 1 (red);
4. tails and section 2 (green).

5.2 Experimental Probability of Simple Events

Grade 7 Module 5 Answer Key Go Math Question 2.
While bowling with friends, Brandy rolls a strike in 6 out of 10 frames. What is the experimental probability that Brandy will roll a strike in the first frame of the next game?
Answer:

The experimental probability that Brandy will roll a strike in the first frame of the next game is \(\frac{3}{5}\)

Question 3.
Ben is greeting customers at a music store. Of the first 20 people he sees enter the store, 13 are wearing jackets and 7 are not. What is the experimental probability that the next person to enter the store will be wearing a jacket?
Answer:

The experimental probability that the next person to enter the store will be wearing a jacket is \(\frac{13}{20}\).

5.3 Experimental Probability of Compound Events

Question 4.
Auden rolled two number cubes and recorded the results.

What is the experimental probability that the sum of the next two numbers rolled is more than 5?
Answer:
From the table we see that the number of times of event occurs is 3 (roll#2, roll#6, and roll#7).
P(the sum > 5) =
= \(\frac{3}{7}\)
The experimental probability that the sum of the next two numbers rolled is more than 5 is \(\frac{3}{7}\).

5.4 Making Predictions with Experimental Probability

Module 5 Test Answers Go Math Grade 7 Quiz Question 5.
A player on a school baseball team reaches first base \(\frac{3}{10}\) of the time he is at bat. Out of 80 times at bat, about how many times would you predict he will reach first base?
Answer:
Use a percent equation.
Find \(\frac{3}{10}\) (30%) of 80.
Write 30% as a fraction. The percent equation will be
x = \(\frac{3}{100}\) ∙ 80 Write fraction as decimal.
= 0.3 ∙ 80 Multiply.
= 24
About 24 times player will reach first base.

Essential Question

Question 6.
How is experimental probability used to make predictions?
Answer:
Experimental probability is comparing the number of times the event occurs to the total number of trials Based on its name, experimentaL, this is the actual result of an experiment done. This is often used for small events which could be done through various trials. One example is when you want to determine the probability of getting a head or a tail when you flip a coin. In experimental probability, you could conduct a trial of flipping a coin 50 times.

The result will be the experimental probability.

Texas Go Math Grade 7 Module 5 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
A frozen yogurt shop offers scoops in cake cones, waffle cones, or cups. You can get vanilla, chocolate, strawberry, pistachio, or coffee-flavored frozen yogurt. If you order a single scoop, how many outcomes are in the sample space?
(A) 3
(B) 5
(C) 8
(D) 15
Answer:
(D) 15

Explanation:
There are 5 flavors of yogurt (vanilla, chocolate, strawberry, pistachio, coffee).
There are 3 ways to serve yogurt (cake cones, waffle cones, cups).
Hence, the sample space is 3 × 5 = 15.

Probability of Compound Events Quiz Grade 7 Question 2.
A bag contains 7 purple beads, 4 blue, beads, and 4 pink beads. What is the probability of not drawing a pink bead?
(A) \(\frac{4}{15}\)
(B) \(\frac{7}{15}\)
(C) \(\frac{8}{15}\)
(D) \(\frac{11}{15}\)
Answer:
(D) \(\frac{11}{15}\)

Explanation:
P(pink bead) =
P(pink bead) = \(\frac{4}{15}\)
The sum of the probabilities of an event and its complement equals 1
P(event) + P(complement) = 1
P(pink bead) + P(not pink bead) = 1 Substitute \(\frac{4}{15}\) for P(pink bead)
\(\frac{4}{15}\) + P(not pink bead) = 1 Subtract \(\frac{4}{15}\) from both sides
P(not pink bead) = 1 – \(\frac{4}{15}\)
P(not pink bead) = \(\frac{11}{15}\)

Question 3.
During the month of June, Ava kept track of the number of days she saw birds in her garden. She saw birds on 18 days of the month. What is the experimental probability that she will see birds in her garden on July 1?
(A) \(\frac{1}{18}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
Answer:
(D) \(\frac{3}{5}\)

Explanation:
P (Ava saw birds) = = \(\frac{18}{30}\) = \(\frac{3}{5}\)
The probability that she will see birds on July 1 is \(\frac{3}{5}\)

Go Math Module 5 Quiz Answers Grade 7 Question 4.
A rectangle has a width of 4 inches and a length of 6 inches. A similar rectangle has a width of 12 inches. What is the length of a similar rectangle?
(A) 8 inches
(B) 12 inches
(C) 14 inches
(D) 18 inches
Answer:
(D) 18 inches

Explanation:
The rectangles are similar, hence, they have corresponding sides.
Write the proportion of the corresponding sides.

The Length of rectangle is 18 inches.

Question 5.
The experimental probability of hearing thunder on any given day in Ohio is 30%. Out of 600 days, on about how many days can Ohioans expect to hear thunder?
(A) 90 days
(B) 180 days
(C) 210 days
(D) 420 days
Answer:
(B) 180 days

Explanation:
Find 30% of 600.
Write 30% as a fraction. The percent equation will. be
x = \(\frac{30}{100}\) ∙ 600 Write fraction as decimal
= 0.3 ∙ 600 Multiply.
= 180
Ohioans can expect to hear thunder about 180 days out of 600 days

Go Math Grade 7 Module 5 Answer Key Question 6.
Isidro tossed two coins several times and then recorded the results in the table below.

What is the experimental probability that both coins will land on the same side on Isidro’s next toss?
(A) \(\frac{1}{5}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{3}{5}\)
(D) \(\frac{4}{5}\)
Answer:
(B) \(\frac{2}{5}\)

Explanation:
Total number of trials is 5: HT, TT, TH, HT, HH
N umber of time he tossed a same side on both coins is 2: TT, HH
Therefore, experimental probability that both of coins Land on same side in next toss is:
P (on both coins are same side) = = \(\frac{2}{5}\)
Experimental probability that both of coins land on same side in next toss is \(\frac{2}{5}\)

Gridded Response

Module 5 Grade 7 Answer Key Go Math Question 7.
Magdalena had a spinner that was evenly divided into sections of red, blue, and green. She spun the spinner and tossed a coin several times. The table below shows the results.

Given the results, what is the experimental probability of spinning blue? Write an answer as a decimal.

Answer:
Total number of trials is 5.
The number of spinning blue is 2: blue-T, blue-H

The experimental probability of spinning blue is 0.4.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 4.4 Answer Key Ratios and Pi.

Texas Go Math Grade 7 Lesson 4.4 Answer Key Ratios and Pi

Texas Go Math Grade 7 Lesson 4.4 Explore Activity Answer Key

Exploring Circumference
A circle is a set of points in a plane that are a fixed distance from the center.

A radius is a line segment with one endpoint at the center of the circle and the other endpoint on the circle. The length of a radius is called the radius of the circle.

A diameter of a circle is a line segment that passes through the center of the circle and whose endpoints lie on the circle. The length of the diameter is twice the length of the radius. The length of a diameter is called the diameter of the circle.

The circumference of a circle is the distance around the circle.

A. Use a measuring tape to find the circumference of five circular objects. Then measure the distance across each item to find its diameter. Record the measurements of each object in the table below.

B. Divide the circumference of each object by its diameter. Round your answer to the nearest hundredth.

Reflect

Question 1.
Make a Conjecture Describe what you notice about the ratio \(\frac{C}{d}\) in your table.
Answer:
All, ratios when calculated are approximately 3.14.

Question 2.
In every circle, what is the ratio of the radius to the diameter? What is the ratio of the circumference to the diameter?
Answer:
In every circle, the ratio of the radius to the diameter is 1 : 2
The ratio of the circumference to the diameter is π

Go Math Lesson 4.4 7th Grade Answer Key Question 3.
Draw Conclusions Based on the ratios of the radius to diameter and circumference to diameter, are all circles proportional? Is the ratio of circumference to the diameter of any circle the same for all circles?
Answer:
Yes, all circles are proportional based on the ratios of the radius to diameter and circumference to diameter.
The ratio of circumference to the diameter of any circle will be equal to π and the same for all circles.

Your Turn

Question 4.
Determine if the radius and diameter of the two circles are proportional.

Answer:
Find the radius and diameter of each circle
First circle:
r₁ = 8 ft
d₁ = 2 × r= 2 × 8= 16 ft
Second Circle:
d₂ = 9 ft
r₂ = d ÷ 2 = 9 ÷ 2 = 4.5 ft
Set up a proportion using the corresponding lengths of the radius and diameter

The radius and diameter of the two circles are proportional.

Go Math Answer Key Grade 7 Lesson 4.4 Question 5.
The circumference of the larger circle is approximately 44 meters and the circumference of the smaller circle is approximately 11 meters. Use a proportion to find the approximate
diameter of the smaller circle.__________

Answer:
C₂ = 44 m
C₁ = 11 m
d₂ = 14 m
Set up a proportion using the given information for the larger and smaller circles.

Texas Go Math Grade 7 Lesson 4.4  Guided Practice Answer Key

Fill in the blanks. (Explore Activity)

Question 1.
Vocabulary In any circle, the ratio of the ___________ to the diameter is pi.
Answer:
In any circle, the ratio of the circumference to the diameter is pi.

Question 2.
Vocabulary In any circle, the ratio of the ____________ radius is 2.
Answer:
In any circle, the ratio of the diameter to the radius is 2.

Question 3.
You can use the decimal number ________ or the fraction ______ as an approximation for pi.
Answer:
You can use the decimal number 3.14 or the fraction \(\frac{C}{d}\) as an approximation for pi.

Circles and Ratios 7th Grade Go Math Question 4.
Determine if the radius and diameter of a circle with a diameter of 100 mm and a circle with a diameter of 10 mm are proportional. (Example 1)
Answer:
Find the radius and diameter of each circle.
First circle:
d₁ = 100 mm
r₁ = d ÷ 2 = 100 ÷ 2 = 50 mm
Second circle:
d₂ = 10 mm
r₂ = d ÷ 2 = 10 ÷ 2 = 5 mm
Set up a proportion using the corresponding Lengths of the radius and diameter.
\(\frac{50}{5} \stackrel{?}{=} \frac{100}{10}\)
10 = 10
The radius and diameter of the two circles are proportional.

Question 5.
Is the circle represented by a penny similar to the one of a quarter? Explain. (Example 1)
Answer:
Every circle is proportional to each other. Thus, the circle represented by a penny is similar to the one of a quarter.

Question 6.
The circumference of the larger circle is about 18.8 centimeters and the circumference of the smaller circle is about 6.3 centimeters Find the approximate diameter of the smaller circle. (Example 2)

Answer:
C₂ = 18.8 cm
C₁ = 6.3 cm
d₂ = 6 m
Set up a proportion using the given information for the larger and smaller circle.
\(\frac{C_{1}}{d_{1}}=\frac{C_{2}}{d_{2}}\)
\(\frac{6.3}{d_{1}}=\frac{18.8}{6}\)
18.8d₁ = 6.3 × 6
18.8d₁ = 37.8
d₁ = 37.8 ÷ 18.8
d₁ ≈ 2 cm

Essential Question Check-In

Question 7.
What is the result of dividing the distance around a circle by the distance across the same circle? What number can you use as an approximate value for this ratio?
Answer:
The result of dividing the distance around a circle by the distance across the same circle is π(pi).
We can use the decimal number 3.14 to approximate the value for π.

Texas Go Math Grade 7 Lesson 4.4  Independent Practice Answer Key

Go Math Grade 7 Lesson 4.4 Answer Key Question 8.
Measurement Jillian measured the distance around a small fishpond as 27 yards. Which would be a good estimate for the distance across the pond, 14 yards, 9 yards, or 7 yards? Explain how you decided.
Answer:
The distance around a small fishpond is the circumference(C).
The distance around the pond is the diameter(d).
C = 27 yd
We know that \(\frac{C}{d}\) ≈ 3.14
\(\frac{C}{d}\) ≈ 3.14
\(\frac{27}{d}\) ≈ 3.14
3.14d ≈ 27
d ≈ 27 ÷ 3.14
d ≈ 8.6 yd
A good estimate for the distance across the pond would be 9 yards.

Question 9.
A rotating wind turbine has a diameter of about 185 feet and its circumference is about 580 feet. A smaller model of the turbine has a circumference of about 10 feet. What will the diameter of the model be?

Answer:
d₁ = 185 ft
C₁ = 580 ft
C₂ = 10 ft
Set up a proportion using the given information for the larger and smaller wind turbine

The diameter of the model will be 319 ft

Question 10.
Multistep Andrew has a flying disc with a radius of 10 centimeters. What is the circumference of the disc? (Remember \(\frac{C}{d}\) = 3.14.)
Answer:
r = 10 cm ⇒ d = 20 cm
Now, calcuLate the circumference of the disc using that \(\frac{C}{d}\) = 3.14
\(\frac{C}{d}\) = 3.14
\(\frac{C}{20}\) = 3.14
C = 20 × 3.14
C = 62.8 cm
The circumference of the disk is 62.8 cm.

Go Math Lesson 4.4 7th Grade Ratio of Circumference Question 11.
Mandie wants to put some lace trim around the outside of a round tablecloth she expanded. The original tablecloth had a radius of 2 feet and a circumference of 12.56 feet. If the tablecloth now has a radius of 3 feet, is 15 feet of lace enough? Explain.
Answer:
Calculate the circumference of the new tablecloth to find out if 15 feet of lace is enough.
r₁ = 2 ft ⇒ d₁ = 4 ft
C₁ = 12.56 ft
r₂ = 3 ft ⇒ d₂ = 6 ft

Since the circumference of the new tablecloth is 18.86 ft which is greater than 15 ft, 15 feet of lace is not enough to put around the outside of the new tablecloth.

Question 12.
Marta is making two different charms for a necklace. One charm has a 1 centimeter diameter and a 3.14 centimeter circumference. A similar charm has a diameter of 4 centimeters. What is the circumference?
Answer:
C = 3.14 cm The circumference of charm
d = 1 cm Diameter of charm
C₁ = ? The circumference of simiLar charm
d₁ = 4 cm Diameter of similar charm
Use a proportion to find the value of the circumference of the similar charm.

The circumference of the similar charm is 12.56 cm.

Question 13.
Randy is putting bricks around the outside of his round flower bed to protect the plants.
a. If the diameter of his flower bed is 100 inches, what is the distance around the garden? (Remember \(\frac{C}{d}\) = 3.14.)
Answer:
To find the distance around the garden, we have to find the circumference of the round flower bed
Use the formula for circumference of the circle
C = π(d) Substitute 100 for d and 3.14 for π.
C ≈ 3.14 ∙ 100
C ≈ 314
The distance around the garden is about 314 in.

b. If the curved bricks he wants to buy are each half a foot long, how many will he need to put around the outside of the garden? Explain.
Answer:
First, convert inches to feet
1 in. = 0.083 ft
C ≈ 314 ∙ 0.83 = 26.26ft
The circumference of the flower bed is 26.26 ft
One brick is half a feet Long, se we have to divide C by \(\frac{1}{2}\) to find how many bricks will he need to buy.
26.06 ÷ \(\frac{1}{2}\) = 26.06 ∙ 2 = 52.12
Randy needs 52.12 bricks to put around the outside of the garden.

c. If each brick costs $0.68, and he can only buy whole bricks, how much will it cost him to get the material to put around the outside of his garden?
Answer:
He needs 52.12 bricks, so he must buy 53 whole bricks.
One brick costs $0.68. therefore
The cost of whole material 0.68 ∙ 53 = 36.04
The material will cost him $36.04.

d. If his mother decides he can only have half of that diameter for his flower bed, how will the cost of the bricks be affected? Explain?
Answer:
The circumference of the flower bed with half of the diameter is half of the circumference with the whole diameter.
Hence, he will need half of the number of bricks, so the cost of the bricks will be halved.
\(\frac{36.04}{2}\) = $18.02

Texas Go Math Grade 7 Answer Key Pdf Lesson 4.4 Question 14.
Your grandmother is teaching you how to make a homemade pie. The pie pan has a diameter of 9 inches.
a. If she asks you to cut a strip of pie crust long enough to go around the outside of the pan, how long does it need to be? (Remember \(\frac{C}{d}\) = 3.14.)
Answer:
C = ? The circumference of pan
d = 9 in Diameter of pan
Use a equation \(\frac{C}{d}\) = 3.14 to find a value of circumference. Substitute value for diameter.
\(\frac{C}{9}\) = 3.14
Multiply both sides by 9.
9 ∙ \(\frac{C}{9}\) = 9 ∙ 3.14
C = 28.26
It need to be 28.26 inches long.

b. If another pie pan is 8 inches across the diameter, how long does that piece of crust need to be?
Answer:
C = ? The circumference of pan
d = 8 + 9 = 17 in Diameter of pan
Use a equation \(\frac{C}{d}\) = 3.14 to find a value of circumference. Substitute value for diameter.
\(\frac{C}{17}\) = 3.14
Multiply both sides by 17.
17 ∙ \(\frac{C}{17}\) = 17 ∙ 3.14
C = 53.38
It need to be 53.38 inches Long.

H.O.T. Focus on Higher Order Thinking

Question 15.
Make a Conjecture You know that all squares are similar and all circles are similar. An equilateral triangle has 3 equal sides and 3 angles of 60 degrees each. Are all equilateral triangles similar?
Answer:
Yes they are, because the corresponding angles of all equilateral triangles are equal, and corresponding sides are proportional.

Question 16.
Multiple Representations You know three different number representations for pi that you can use to approximate the answer to a problem. Describe a situation when you might choose to use \(\frac{22}{7}\).
Answer:
You might choose to use \(\frac{22}{7}\) when you have a diameter whose value is 7.
For example:
Find the circumference of the circle if the diameter is 7 centimeters.
C = ? The circumference of a circle
d = 7 cm Diameter of circle
Use the equation \(\frac{C}{d}\) = \(\frac{22}{7}\) to find a value of circumference Substitute value for diameter.
\(\frac{C}{7}\) = \(\frac{22}{7}\)
Therefore C = 22 centimeters.

Texas Go Math Grade 7 Lesson 4.4 Answer Key Question 17.
Represent Real-World Problems Describe an example in your daily life where you might be able to measure around something but could not measure across it.
Answer:
One example is that when you want to measure the circumference of a post. We could use a measuring tool to determine the circumference of the post by simply going around the post. However, if we want to determine the area of the post, it is somewhat impossible. First, the post was already attached to the ground and depending on the height of the post, we could not check the diameter or radius of the post.

Another one is measuring the circumference and area of a tree. We could find the circumference of a tree by simply going around the tree. But if you want to measure the area of a tree, it is impossible unless you cut the tree to determine the diameter or radius of the tree.

You can measure the circumference of a post or tree but you can’t measure their area.

Question 18.
Critical Thinking Every morning Jesse runs 3 laps on a circular track. One morning the track is closed, but the straight path from one side of the track to the other is open. How many times should Jesse run across the path if he wants to run his usual distance? Explain your answer.
Answer:
As Jesse runs 3 Laps on a circular track, he crosses 3 ∙ C where C represents the circumference of the track.
The straight path in this case represents the diameter d of this circular track.
Therefore, if you want to calculate how many times he must run across the path to run his usual distance, use the formula \(\frac{C}{d}\) = 3.14 to find C.
C = 3.14 ∙ d
He crosses the circular track 3 times, therefore, 3 ∙ C = 9.42 ∙ d
He must run across the path to run his usual distance of 9.42 times.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 5 Answer Key Experimental Probability.

Texas Go Math Grade 7 Module 5 Answer Key Experimental Probability

Texas Go Math Grade 7 Module 5 Are You Ready? Answer Key

Write each fraction in simplest form.

Question 1.
\(\frac{6}{10}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
6 : 1, 2, 3, 6
10 : 1, 2, 5, 10
The greatest common factor is 2.
Divide the numerator and denominator by the greatest common factor.
\(\frac{6 \div 2}{10 \div 2}\) = \(\frac{3}{5}\)

Go Math Grade 7 Module 5 Answer Key Question 2.
\(\frac{9}{15}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
9 : 1, 3, 9
15 : 1, 3, 5, 15
The greatest common factor is 3.
Divide the numerator and denominator by the greatest common factor.
\(\frac{9 \div 3}{15 \div 34}\) = \(\frac{3}{5}\)

Question 3.
\(\frac{16}{24}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
16 : 1, 2, 4, 8, 16
24 : 1, 2, 3, 4, 6, 8, 12, 24
The greatest common factor is 8.
Divide the numerator and denominator by the greatest common factor.
\(\frac{16 \div 8}{24 \div 8}\) = \(\frac{2}{3}\)

Question 4.
\(\frac{9}{36}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
9 : 1, 3, 9
36 : 1, 2, 3, 4, 6, 9, 12, 18, 36
The greatest common factor is 9.
Divide the numerator and denominator by the greatest common factor.
\(\frac{9 \div 9}{36 \div 9}\) = \(\frac{1}{4}\)

Question 5.
\(\frac{45}{54}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
45 : 1, 3, 5, 9, 15, 45
54 : 1, 2, 3, 6, 9, 18, 27, 54
The greatest common factor is 9.
Divide the numerator and denominator by the greatest common factor.
\(\frac{45 \div 9}{54 \div 9}\) = \(\frac{5}{6}\)

Go Math Grade 7 Module 5 Topic A Quiz Answer Key Question 6.
\(\frac{30}{42}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
30 : 1, 2, 3, 5, 6, 10, 15, 30
42 : 1, 2, 3, 6, 7, 14, 21, 42
The greatest common factor is 6.
Divide the numerator and denominator by the greatest common factor.
\(\frac{30 \div 6}{42 \div 6}\) = \(\frac{5}{7}\)

Question 7.
\(\frac{36}{60}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
36 : 1, 2, 3, 4, 6, 9, 12, 18, 36
60 : 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
The greatest common factor is 12.
Divide the numerator and denominator by the greatest common factor.
\(\frac{36 \div 12}{60 \div 12}\) = \(\frac{3}{5}\)

Question 8.
\(\frac{14}{42}\) ____________
Answer:
Find the greatest common factor of the numerator and denominator.
List factors of the numerator and denominator.
14 : 1, 2, 7, 14
42 : 1, 2, 3, 6, 7, 14, 21, 42
The greatest common factor is 14.
Divide the numerator and denominator by the greatest common factor.
\(\frac{14 \div 14}{42 \div 14}\) = \(\frac{1}{3}\)

Write each fraction as a decimal.

Question 9.
\(\frac{3}{4}\) ____________
Answer:
Divide 3 by 4. Write a decimal point and insert extra zeros in the dividend.

The result is 0.75.

Grade 7 Module 5 Review Answer Key Go Math Question 10.
\(\frac{7}{8}\) ____________
Answer:
Divide 7 by 8. Write a decimal point and insert extra zeros in the dividend.

The result is 0.875.

Question 11.
\(\frac{3}{20}\) ____________
Answer:
Divide 3 by 20. Write a decimal point and insert extra zeros in the dividend.

The result is 0.15.

Question 12.
\(\frac{19}{50}\) ____________
Answer:
Divide 19 by 50. Write a decimal point and insert extra zeros in the dividend.

The result is 0.38.

Write each percent as a decimal.

Question 13.
67% ______
Answer:
Write the 67% as the sum of the 1 whole and the percent remainder.
= 100% – 33% Write the percents as a fractions.
= \(\frac{100}{100}\) – \(\frac{33}{100}\) Divide.
= 1 – 0.33 Subtract
= 0.67

Question 14.
31% ______
Answer:
Write the 31% as the sum of the 1 whole and the percent remainder.
= 100% – 69% Write the percents as a fraction.
= \(\frac{100}{100}\) – \(\frac{69}{100}\) Divide.
= 1 – 0.69 Subtract
= 0.31

Question 15.
7% ____________
Answer:
Write the 7% as the sum of the 1 whole and the percent remainder.
= 100% – 93% Write the percents as a fraction.
= \(\frac{100}{100}\) – \(\frac{93}{100}\) Divide.
= 1 – 0.93 Subtract
= 0.7

Question 16.
54% ____________
Answer:
Write the 54% as the sum of the 1 whole and the percent remainder.
= 100% – 46% Write the percent as a fraction.
= \(\frac{100}{100}\) – \(\frac{46}{100}\) Divide.
= 1 – 0.46 Subtract
= 0.54

Write each decimal as a percent.

Question 17.
0.13 ____________
Answer:
Multiply 0.13 by 1. Any number will be the same when multiplied by 1.
0.13 ∙ 1 = 0.13 ∙ \(\frac{100}{100}\) Write 1 as a fraction with the denominator 100.
= \(\frac{13}{100}\) Multiply Write fraction as a percent
= 13%

Question 18.
0.55 ____________
Answer:
Multiply 0.55 by 1. Any number will be the same when multiplied by 1.
0.55 ∙ 1 = 0.55 ∙ \(\frac{100}{100}\) Write 1 as a fraction with the denominator 100.
= \(\frac{55}{100}\) Multiply Write fraction as a percent
= 55%

Question 19.
0.08 ____________
Answer:
Multiply 0.08 by 1. Any number will be the same when multiplied by 1.
0.08 ∙ 1 = 0.08 ∙ \(\frac{100}{100}\) Write 1 as a fraction with the denominator 100.
= \(\frac{8}{100}\) Multiply Write fraction as a percent
= 8%

Grade 7 Module 5 Answer Key Go Math Question 20.
1.16 ____________
Answer:
Multiply 1.16 by 1. Any number will be the same when multiplied by 1.
1.16 ∙ 1 = 1.16 ∙ \(\frac{100}{100}\) Write 1 as a fraction with the denominator 100.
= \(\frac{116}{100}\) Multiply Write fraction as a percent
= 116%

Texas Go Math Grade 7 Module 5 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the graphic. You can put more than one word in each box.

Understand Vocabulary

Match the term on the left to the definition on the right.

Answer:
1 – A. The probability – measures the likelihood that the event will occur.
3 – C. Each observation of an experiment is a trial.
2 – B. A set of one or more outcomes is an event

Active Reading
Pyramid Before beginning the module, create a rectangular pyramid to help you organize what you learn. Label each side with one of the lesson titles from this module. As you study each lesson, write important ideas, such as vocabulary, properties, and formulas, on the appropriate side.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 6.3 Answer Key Making Predictions with Theoretical Probability.

Texas Go Math Grade 7 Lesson 6.3 Answer Key Making Predictions with Theoretical Probability

Your Turn

Question 1.
Predict how many times you will roll a number less than 5 if you roll a standard number cube 250 times.
Answer:
The probability of rolling number less than 5 is \(\frac{4}{6}\) = \(\frac{2}{3}\)
Use proportion.
The probability of rolling number less than 5 is equal to what number “x divided by 150?

To roll a number less than 5, you should roll a cube 167 times.

Texas Go Math Grade 7 Solutions Lesson 6.3 Answer Key Question 2.
You flip a fair coin 18 times. About how many times would you expect heads to appear?
Answer:
The probability of flipping a coin and getting a head is \(\frac{1}{2}\).
Use proportion.
The probability to flip a coin and get a head is equal to what number ‘x’ divided by 18?

To get a head, you should flip a coin 9 times.

Question 3.
A bag of marbles contains 8 red marbles, 4 blue marbles, and 5 white marbles. Tom picks a marble at random, Is it more likely that he picks a red marble or a marble of another color?
Answer:
P (picking red marble) = \(\frac{8}{17}\)
P (picking a marble with another color) = 1 – P (picking red marble) = 1 – \(\frac{8}{17}\) = \(\frac{9}{17}\)
The probability to pick red marble is less than a probability to pick a marbLe of another color because \(\frac{8}{17}\) < \(\frac{9}{17}\).

Question 4.
At a fundraiser, a school group charges $6 for tickets for a “grab bag.” You choose one bill at random from a bag that contains 40 $1 bills, 20 $5 bills, 5 $10 bills, 5 $20 bills, and 1 $100 bill. Is it likely that you will win enough to pay for your ticket? Justify your answer.
Answer:
Total number of possible outcomes is: 40 + 20 + 5 + 5 + 1 = 71.
Number of $10, $20 and $100 bills is 5 + 5 + 1 = 11.\$
P (win enough to pay ticket) = = \(\frac{11}{71}\)
1 – P (win enough to pay ticket) = 1 – = 1 – \(\frac{11}{71}\) = \(\frac{60}{71}\)
It is more likely not to win enough to pay a ticket.

Texas Go Math Grade 7 Lesson 6.3 Guided Practice Answer Key

Question 1.
Bob works at a construction company. He has an equally likely chance of being assigned to work different crews every day. He can be assigned to work on crews building apartments, condominiums, or houses. If he works 18 days a month, about how many times should he expect to be assigned to the house crew? (Example 1)
STEP 1 Find the probabilities of beg assigned to each crew.

The probability of being assigned to the house crew is _________.

STEP 2: Set up and solve a proportion.

Bob can expect to be assigned to the house crew about _______ times out of 18.
Answer:

The probability of being assigned to the house crew is \(\frac{1}{3}\)
Bob can expect to be assigned to the house crew about 6 times out of 18.

Making Predictions with Theoretical Probability Lesson 6.3 Answer Key Question 2.
During a raffle drawing, half of the ticket holders will receive a prize. The winners are equally likely to win one of three prizes: a book, a gift certificate to a restaurant, or a movie ticket. If there are 300 ticket holders, predict the number of people who will win a movie ticket. (Example 1)
Answer:
P(win) = \(\frac{1}{2}\)
P (win a movie ticket) =
Use a proportion.
\(\frac{1}{3}\) = \(\frac{x}{150}\)
x = 50
About 50 peoples will win a movie tickets.

Question 3.
In Mr. Jawarani’s first period math class, there are 9 students with hazel eyes, 10 students with brown eyes, 7 students with blue eyes, and 2 students with green eyes. Mr. Jawarani picks a student at random. Which color eyes is the student most likely to have? Explain. (Example 2)

Answer:
Total number of possible outcomes is: 9 + 10 + 7 + 2 = 28.

It is more likely to pick a student with brown eyes.

Essential Question Check-In

Question 4.
How do you make predictions using theoretical probability?
Answer:
Because theoretical and experimental probabilities are ratios, we can use proportions with probabilities to make predictions. Therefore using an equation where the theoretical probability is equal to number of x divided by a total number of outcomes, we get a prediction for some event, which is number x.

Texas Go Math Grade 7 Lesson 6.3 Independent Practice Answer Key

Question 5.
A bag contains 6 red marbles, 2 white marbles, and 1 gray marble. You randomly pick out a marble, record its color, and put it back in the bag. You repeat this process 45 times. How many white or gray marbles do you expect to get?
Answer:
Total. number of possibIe outcomes is: 6 + 2 + 1 = 9

You can expect about 15 grey or white marbles.

Lesson 6.3 Answer Key 7th Grade Go Math Question 6.
Using the blank circle below, draw a spinner with 8 equal sections and 3 colors—red, green, and yellow. The spinner should be such that you are equally likely to land on green or yellow, but more likely to land on red than either on green or yellow.

Answer:


P (land on red) = \(\frac{6}{8}\)
P (land on yellow) = \(\frac{1}{8}\)
P (land on green) = \(\frac{1}{8}\)

Use the following for Exercises 7-9.

In a standard 52-card deck, half of the cards are red and half are black. The 52 cards are divided evenly into 4 suits: spades, hearts, diamonds, and clubs. Each suit has three face cards (jack, queen, king), and an ace. Each suit also has 9 cards numbered from 2 to 10.

Question 7.
Dawn draws 1 card, replaces it, and draws another card. Is it more likely that she draws 2 red cards or 2 face cards?
Answer:
Total number of possible outcomes is:
red-black, red-red, black-black, black-red = 4
P (draw two red cards) =

Total number of possible outcomes is:
52 possibilities ∙ 52 possibilities = 52 ∙ 52 = 2704
P (draw two face cards) =
The probability to pick two red cards is more likely than the probability to pick two face cards.

Question 8.
Luis draws 1 card from a deck, 39 times. Predict how many times he draws an ace.
Answer:

You can predict that an ace will be drawn about 3 times.

Go Math Answer Key Grade 7 Probability Lesson 6.3 Question 9.
Suppose a solitaire player has played 1,000 games. Predict how many times the player turned over a red card as the first card.
Answer:

You can predict that the player will, turn over the red card as the first card about 500 times.

Question 10.
John and O’Neal are playing a board game in which they roll two number cubes. John needs to get a sum of 8 on the number cubes to win. O’Neal needs a sum of 11. If they take turns rolling the number cube, who is more likely to win? Explain.
Answer:

It is more likely John to win.

Question 11.
Every day, Navya’s teacher randomly picks a number from 1 to 20 to be the number of the day. The number of the day can be repeated. There are 180 days in the school year. Predict how many days the number of the day will be greater than 15.
Answer:

You can expect the number of day will be greater than 5 about 45 days.

Go Math 7th Grade Lesson 6.3 Theoretical Probability Answer Key Question 12.
Eben rolls two standard number cubes 36 times. Predict how many times he will roll a sum of 4.
Answer:

About 45 times will roll a sum of 4.

Question 13.
Communicate Mathematical Ideas Can you always show that a prediction based on theoretical probability is true by performing the event often enough? If so, explain why. If not, describe a situation that justifies your response.
Answer:
Answer is no.
Example:
A coin is flipped 16 times and head is obtained 7 times. Find a experimental. probability to flip a head.

Question 14.
Represent Real-World Problems Give a real-world example of an experiment in which all of the outcomes are not equally likely. Can you make a prediction for this experiment, using theoretical probability?
Answer:
What is the probability of rolling a 2 or 5 on a 6-sided cube with one of next numbers on each side: 1, 2, 2, 3, 4,5?
How 2 appear on two sides of the cube, and 5 only on one, it is more likely that we will roll a 2.
Using the theoretical probability we will show the same.

H.O.T. Focus on Higher Order Thinking

Question 15.
Critical Thinking Pierre asks Sherry a question involving the theoretical probability of a compound event in which you flip a coin and draw a marble from a bag of marbles. The bag of marbles contains 3 white marbles, 8 green marbles, and 9 black marbles. Sherry’s answer, which is correct, is \(\frac{12}{40}\). What was Pierre’s question?
Answer:
Pierre’s question was: What is the probability to pick white or black marble and flip a head?

Go Math Grade 7 Lesson 6.3 Making Predictions Question 16.
Make a Prediction Horace is going to roll a standard number cube and flip a coin. He wonders if it is more likely that he rolls a 5 and the coin lands on heads, or that he rolls a 5 or the coin lands on heads. Which event do you think is more likely to happen? Find the probability of both events to justify or reject your initial prediction.
Answer:
Total number of possible outcomes for rolling 5 and flip a head is: 6 possibilities ∙ 2 possibilities 6 ∙ 2 = 12
Number of rolling a 5 and flip a head is only one.

Number of head or rolling 5 is: head-1, head-2, head-3, head-4, head-6 and head-5 and letter-5 = 7

To flip a head or roll a 5 is more likely to happen.

Question 17.
Communicate Mathematical Ideas Cecil solved a theoretical prediction problem and got this answer: “The spinner will land on the red section 4.5 times.” Is it possible to have a prediction that is not a whole number? If so, give an example.
Answer:
Use spinner with eight equal sections with 4 red, 2 green, and 2 blue sections
Cecil spins a spinner shown. Predict how many time she spins on red if she spins 9 times

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Unit 2 Study Guide Review Answer Key.

Texas Go Math Grade 7 Unit 2 Study Guide Review Answer Key

Texas Go Math Grade 7 Unit 2 Exercises Answer Key

Module 2 Rates and Proportionality

Question 1.
Steve uses \(\frac{8}{9}\) gallon of paint to paint 4 identical birdhouses. How many gallons of paint does he use for each birdhouse? (Lesson 2.1)
Answer:
For 4 identical birdhouses Steve uses \(\frac{8}{9}\) gallons of paint.
To find how many gallons he needs for one birdhouse, we have to divide \(\frac{8}{9}\) by 4.
\(\frac{8}{9}\) ÷ 4 = \(\frac{8}{9}\) ∙ \(\frac{1}{4}\) Multiply by the reciprocaL of the divisor
= \(\frac{8 \cdot 1}{9 \cdot 4}\)
= \(\frac{8}{36}\)
= \(\frac{4 \cdot 2}{4 \cdot 9}\)
= \(\frac{2}{9}\)
He uses \(\frac{2}{9}\) gallons of paint for each birdhouse.

Grade 7 Mathematics Unit 2 Answer Key Question 2.
Ron walks 0.5 miles on the track in 10 minutes. Stevie walks 0.25 miles on the track in 6 minutes. Find the unit rate for each walker in miles per hour. Who is the faster walker? (Lesson 2.1)
Answer:
To find how much miles Ron waLks for one minute, divide the number of miles by a number of minutes.
\(\frac{0.5}{10}\) = 0.05
For one minute he waLks 0.05 miles.
For one hour he walks 0.05 ∙ 60 = 3 miles.
Therefore, he walks 3 miles per hour.
Repeat procedure with Stevie.
\(\frac{0.25}{6}\) = 0.04
For one hour he walks 0.04 ∙ 60 = 2.4 miles.
Therefore, he walks 2.4 miles per hour.
We get 3 > 2.4, therefore, Ron is the faster walker.

Question 3.
(Lessons 2.2, 2.3) The table below shows the proportional relationship between Juan’s pay and the hours he works. Complete the table. Plot the data and connect the points with a line. (Lessons 2.2, 2.3)

Answer:
First, find the constant of proportionality.
Let y be pay ($) and x be hours worked.
The constant of proportionality = \(\frac{y}{x}\) = \(\frac{40}{2}\) = 20
Hence, for I hour of work he earns $20.
To find how many hours he needs to work for $80. we divide 80 by the constant of proportionality.
\(\frac{80}{20}\) = 4
For 1 hour he earns $20, so for 5 hours he earns 5 ∙ 20 = $100.
For 1 hour he earns $20. so for 6 hours he earns 6 ∙ 20 = $120.

Module 3 Proportions and Percent

Convert each measurement. (Lesson 3.1)

Texas Go Math Study Guide 7th Grade Unit 2 Question 1.
7 centimeters ≈ __________ inches
Answer:
As we know 1 inch ≈ 2.54 centimeters.
Divide both sides by 2.54.
You get:
≈ 1 cm
To calculate for 7 centimeters, multiply both sides by 7.
7 ∙ \(\frac{1 \mathrm{~ in}}{2.54}\) ≈ 7 ∙ 1 cm
7 0.4 in ≈ 7 cm
2.8 in ≈ 7 cm

Question 2.
10 pounds ≈ __________ kilograms
Answer:
1 pound ≈ 0.45 kg
10 pound ≈ 4.5 kg

Question 3.
24 kilometers ≈ ___________ miles
Answer:
As we know 1 mile ≈ 1.61 kilometers
Divide both sides by 1.61.
You get:
\(\frac{1 \mathrm{~ mi}}{1.61}\) ≈ 1 km
To calculate for 24 kilometers. multipLy both sides by 24.
24 ∙ \(\frac{1 \mathrm{~ mi}}{1.61}\) ≈ 24. 1 km
24 ∙ 0.62 mi ≈ 24 km
14.9 mi ≈ 24 km

Question 4.
12 quarts ≈ _____________ liters
Answer:
1 quart ≈ 0.95 l
12 quart ≈ 11.4 l

Question 5.
Michelle purchased 25 audio files in January. In February she purchased 40 audio files. Find the percent increase. (Lesson 3.2)
Answer:
First we find amount of change.
Amount of Change = Purchased audio files in February – Purchased audio files in January
= 40 – 25
= 15
Now, we have all, necessary data to find percent change.

= \(\frac{15}{25}\)
= \(\frac{3}{5}\)
= 0.6 = 60%

Texas Go Math Unit 2 Assessment Answer Key Question 6.
Sam’s dog weighs 72 pounds. The vet suggests that for the dog’s health, its weight should decrease by 12.5 . percent. According to the vet, what is a healthy weight for the dog? (Lesson 3.2)
Answer:
Find 12.5 percent od 72 and subtract it from 72
Write percent as fractor.
\(\frac{12.5}{100}\) ∙ 72 = 0.125 ∙ 72 = 9
The healthy weight for the dog is 72 – 9 = 63 pounds.

Question 7.
The original price of a barbecue grill is $79.50. The grill is marked down 15%. What is the sale price of the grill? (Lesson 3.3)
Answer:
Original price is $79.50.
Find 15% percent of original price.
79.50 ∙ \(\frac{15}{100}\) = 79.50 ∙ 0.15 = 11.9
Subtract the result from original price.
79.50 – 11.9 = $67.6
Therefore, the sale price of grill is $67.6.

Question 8.
A sporting goods store marks up the cost s of soccer balls by 250%. Write an expression that represents the retail cost of the soccer balls. The store buys soccer balls for $5.00 each. What is the retail price of the soccer balls? (Lesson 3.3)
Answer:
Store pays $5 for each soccer hail and sells it for 250% of the price she pays for the ball.
Find the 250% of the 5.
Write percent as a fraction.
\(\frac{250}{100}\) ∙ 5 = 2.55
= 12.5
The retail price of the soccer hail is $12.5.

Module 4 Proportionality in Geometry

Question 1.
Are the four-sided shapes similar? Explain. (Lesson 4.1)
Answer:
\centering From the picture we can see that:
m∡M = m∡Q = 80°; ∡M corresponds to ∡Q
m∡O = m∡S = 125°; ∡O corresponds to ∡S
\centering AngLes ∡N and ∡P do not correspond to angles ∡R and ∡T.
\(\overline{M P}\) corresponds to \(\overline{Q T}\)
\(\overline{M N}\) corresponds to \(\overline{Q R}\)
\(\overline{N O}\) corresponds to \(\overline{R S}\)
\(\overline{M P}\) corresponds to \(\overline{Q T}\)
\(\overline{P O}\) corresponds to \(\overline{T S}\)
Is it true that \(\frac{M P}{Q T}=\frac{M N}{Q R}=\frac{N O}{R S}=\frac{P O}{T S}\)?
\(\frac{10}{15}=\frac{6}{9}=\frac{8}{12}=\frac{4}{6.5}\)
We get:
0.66 = 0.66 = 0.66 = 0.61
which is not correct, therefore, these four-sided shapes are not similar

Texas Go Math Grade 7 Solutions Unit 2 Answer Key Question 2.
△JNZ ~ △KOA. Find the unknown measures. (Lesson 4.2)

x = ______________
y = ______________
r = ______________
s = ______________
Answer:
To find y, we use Pythagoras theorem:
a² + b² = c²
where C represents the Length of the hypotenuse and a and b the lengths of the triangLes other two sides.
y² = 8² + 15²
y² = 64 + 225
y² = 289 Root both side
\(\sqrt{y^{2}}=\sqrt{289}\)
y = 17
These two shapes are similar, so the corresponding sides are proportional.
Write the proportion.

The sum of angles in a triangle is 180°.
Hence
62° + 90° + r = 180°
152° + r = 180° Subtract 152° from both sides
r = 180° – 152°
r = 28°

28° + 90° + s = 180°
118° + s = 180° Subtract 118° from both sides.
s = 180° – 118°
s = 62°

Question 3.
In the scale drawing of a park, the scale is 1 cm: 10 m. Find the area of the actual park. (Lesson 4.3)

Answer:
1 centimeter in this drawing equals 10 meters oo the actual park.
Find the length of the actual park.
MuLtiply 3 by 10.
3 ∙ 10 = 30 m
Find a width of actual park.
Multiply 1.5 by 10.
1.5 ∙ 10 = 15
The area of the actual park find from formula for the area of rectangle P = l ∙ w.
P = 30 ∙ 15 = 450 m²

Unit 2 Test Review Math Answers 7th Grade Question 4.
The circumference of the larger circle is 1 5.7 yards. Find the circumference of the smaller circle. (Lesson 4.4)

Answer:
C₁ = the circumference of the larger circle
C₂ = the circumference of the smaller circle
d₁= the diameter of the larger circle
d₂ = the diameter of the smaller circle
Two circles are similar, so the corresponding measures are proportional.
Write the proportion.

The circumference of the smaller circle is 12.56 yards.

Texas Go Math Grade 7 Unit 2 Performance Tasks Answer Key

Question 1.
CAREERS IN MATH Landscape Architect A landscape architect creates a scale drawing of her plans for a garden. She draws the plans on a sheet of paper that measures 8\(\frac{1}{2}\) inches by 11 inches. On the right- hand side of the paper, there is a column 2\(\frac{1}{2}\) inches wide that includes the company name and logo. The drawing itself is 5\(\frac{1}{2}\) inches by 7\(\frac{1}{2}\) inches. The scale of the drawing is 1 inch = 10 feet.

a. The landscape architect wants to make a larger drawing on a sheet of paper that measures 11 inches by 17 inches. The larger drawing should include the same 2\(\frac{1}{2}\) inch column on the side. There should be at least \(\frac{1}{2}\) inch of space on all sides of the drawing. What are the dimensions of the area she can use to make the new drawing?
Answer:
From the 17 inches, the 2\(\frac{1}{2}\) inches removed for the area of the sketch there is 14\(\frac{1}{2}\) Removing the \(\frac{1}{2}\) inch for the border, the dimensions of the area will be 10 inches by 13\(\frac{1}{2}\) inches. The sketch of the dimensions for the 11 inches by 17 inches is:

b. How large can she make the scale drawing without changing any of the proportions? Justify your reasoning.
Answer:
Determine how large the scale drawing can be without changing the proportion. Let x be the length of the other side.
\(\frac{5.5}{7.5}\) = \(\frac{x}{13.5}\) Set up the proportion
74.25 = 7.5x Cross multiply
\(\frac{74.25}{7.5}\) = \(\frac{7.5x}{7.5}\) Divide both sides by 7.5
9.9 = x Simplify

Unit 2 Mid Unit Assessment Answer Key Grade 7 Question 2.
The table below shows how far several animals can travel at their maximum speeds in a given time.

a. Write each animal’s speed as a unit rate in feet per second.
Answer:
Elk travels 33 ft in \(\frac{1}{2}\) seconds.
To find the unit rate in feet per second, divide the distance by time.
33 ÷ \(\frac{1}{2}\) = 33 ∙ 2 = 66
Speed of elk is 66 ft/second.
Giraffe travels 115 ft in 2\(\frac{1}{2}\) seconds.
To find the unit rate in feet per second, divide the distance by time.
115 ÷ 2\(\frac{1}{2}\) = 115 ÷ \(\frac{5}{2}\) = 115 ÷ 2.5 = 46
Speed of giraffe is 46 ft/second.
Zebra travels 117 ft in 2 seconds.
To find the unit rate in feet per second, divide the distance by time.
117 ÷ 2 = 117 ÷ 2 = 58.5
Speed of zebra is 58.5 ft/second.

b. Which animal has the fastest speed?
Answer:
The elk is fastest animal.

c. How many miles could the fastest animal travel in 2 hours if it maintained the speed you calculated in part a? Use the formula d = rt and round your answer to the nearest tenth of a mile. Show your work.
Answer:
First, convert 2 hours to seconds, and 66 ft/second to miles/second
1 foot = 0.00019 miles
1 hours = 3600 seconds
2 hours = 2 3600 = 7200 seconds
66 ft/s = 66 0.00019 = 0.0 12 mi/s
d = r ∙ t Substitute 0.012 for r and 7200 for t.
d = 0.012 ∙ 7200
d = 86.4
The fastest animal, in this case elk, travels 86 miles in 2 hours.

d. The data in the table represents how fast each animal can travel at its maximum speed. Is it reasonable to expect the animal from part b to travel that distance in 2 hours? Explain why or why not.
Answer:
Yes, it is reasonable, because the average top speed of the elk is about 45 miles per hour.

Texas Go Math Grade 7 Unit 2 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
If the relationship between distance y in feet and time x in seconds is proportional, which rate is represented by \(\frac{y}{x}\) = 0.6?
(A) 3 feet in 5 s
(B) 3 feet in 9 s
(C) 10 feet in 6 s
(D) 18 feet in 3 s
Answer:
(A) 3 feet in 5 s

Explanation:
Variable x represents time in seconds.
Variable y represents distance in feets.
We need to get. that x divided by y is 0.6.
check answer under (A) when we substitute value for x and y.

The rate represented by \(\frac{x}{y}\) = 0.6 is under (A) 3 feet in 5 seconds.

Algebra 1 Unit 2 Study Guide Texas Go Math Grade 7 Question 2.
The Baghrams make regular monthly deposits in a savings account. The graph shows the relationship between the number x of months and the amount y in dollars in the account.

What is the equation for the deposit?
(A) \(\frac{y}{x}\) = $25/month
(B) \(\frac{y}{x}\) = $40/month
(C) \(\frac{y}{x}\) = $50/month
(D) \(\frac{y}{x}\) = $75/month
Answer:
(C) \(\frac{y}{x}\) = $50/month

Explanation:
For 1 month, the amount in account is $50.
Hence.
\(\frac{y}{x}\) = \(\frac{50}{1}\) = $50/month

For 2 month, the amount in account is $100.
Hence,
\(\frac{y}{x}\) = \(\frac{100}{2}\) = $50/month

For 3 month, the amount in account is $150.
Hence,
\(\frac{y}{x}\) = \(\frac{150}{3}\) = $50/month
The answer is \(\frac{y}{x}\) = \(\frac{50}{1}\) = $50/month.

Hot Tip! Read graphs and diagrams carefully. Look at the labels for important information.

Question 3.
Rosa’s room is 4 meters wide. Which of these is an equivalent measurement?
(A) 0.28 mile
(B) 4.38 yards
(C) 12.4 feet
(D) 136.2 inches
Answer:
(B) 4.38 yards

Explanation:
\centering As we know:
0.39 = \(\frac{1}{2.54}\) inch ≈ 1 centimeters
3.28 = \(\frac{1}{0.305}\) foot ≈ 1 meters
1.094 = \(\frac{1}{0.914}\) yard ≈ 1 meters
0.62 = \(\frac{1}{1.61}\) mile ≈ 1 kilometers

One meter has 100 centimeters. so. four meters have 400 centimeters.
Multiply 0.39 by 100 to obtain how much inches is in four meters.
0.39 ∙ 400 = 156 inches
Therefore, the answer under (D) is not correct.
To obtain how much feet is in four meters, multiply 3.28 by 4.
3.28 ∙ 4 = 13.12 ft
Therefore, the answer under (C) is not, correct.
To obtain how many yards are in four meters, multiply 1 .094 by 4.
1.091 ∙ 1 = 4.38 yards
Therefore, the answer under (B) is correct.
One kilometer has 1000 meters. so. four kilometers have 4000 meters.
To obtain how many miles is in four meters, multiply 0.62 by 4000.
0.62 ∙ 4000 = 2480 miles
Therefore, the answer under (A) is not correct.

Unit 2 Study Guide Answer Key Grade 7 Go Math Question 4.
What is the decimal form of -4\(\frac{7}{8}\)?
(A) -4.9375
(B) -4.875
(C) -4.75
(D) -4.625
Answer:
(B) -4.875

Explanation:
First, convert whole number 4 to eights and add \(\frac{7}{8}\).
\(-\frac{4 \cdot 8+7}{8}=-\frac{32+7}{8}=-\frac{39}{8}\) = -4.875.

Question 5.
Find the percent change from 72 to 90.
(A) 20% decrease
(B) 20% increase
(C) 25% decrease
(D) 25% increase
Answer:
(D) 25% increase

Explanation:
First we find amount of change.
Amount of Change = Greater Value Lesser Value
= 90 – 72
= 18
Now, we have all necessary data to find percent change.

= \(\frac{18}{72}\)
= \(\frac{1}{4}\)
= 0.25 = 25%

Question 6.
A store had a sale on art supplies. The price p of each item was marked down 60%. Which expression represents the new price?
(A) 0.4p
(B) 0.6p
(C) 1.4p
(D) 1.6p
Answer:
(A) 0.04p

Explanation:
Because the price of each item decreases 60%, the new price of each item is 40% of price p.
40% as a decimal. is 0.4.
Hence, the new price for each item is 0.04p.

Question 7.
Clarke borrows $ 16,000 to buy a car. He pays simple interest at an annual rate of 6% over a period of 3.5 years. How much does he pay altogether?
(A) $18,800
(B) $19,360
(C) $19,920
(D) $20,480
Answer:
(B) $19,360

Explanation:
6 % in decimal amounts 0.06.
We are searching for an amount of simple interest for one year.
Multiply 0.06 by 16, 000.
0.06 ∙ 16000 = $960
To find for 3.5 years, multiply the result by 3.5.
960 ∙ 3.5 = $3,360
Altogether he pay:
16,000 + 3,360 = $19,360

Question 8.
To which set or sets does the number 37 belong?
(A) integers only
(B) rational numbers only
(C) integers and rational numbers only
(D) whole numbers, integers, and rational numbers
Answer:
(D) whole numbers, integers, and rational numbers

Explanation:
Number 37 is whole number, integer and rational number.

Question 9.
The two triangles below are similar. What is the missing length?

(A) 21
(B) 22
(C) 24
(D) 26
Answer:
(A) 21

Explanation:
a₁ = 14
b₁ = 16
a = ?
b = 24
\centering To find the missing Length, use a proportion

Gridded Response

Question 10.
The smaller circle has a diameter that is half the size of the larger circle. What is the missing circumference in centimeters?

Answer:
C₁ = the circumference of the larger circle
C₂ = the circumference of the smaller circle
d₁ = the diameter of the larger circle
d₂ = the diameter of the smaller circle
The diameter of the smaller circle is half the diameter of the larger circle.
Hence
d₁ = 2d₂
Two circles are similar, so the corresponding measures are proportional.
Write the proportion.
\(\frac{C_{1}}{C_{2}}=\frac{d_{1}}{d_{2}}\) Substitute 53.38 for C_1, 2d₂ for d₁.
\(\frac{53.38}{C_{2}}=\frac{d_{2}}{d_{2}}\)
\(\frac{53.38}{C_{2}}\) = 2
C₂ = \(\frac{53.38}{2}\)
C₂ = 26.69
The circumference of the smaller circle is 26.69 cm.

7th Grade Math Study Guide Pdf Unit 2 Test Answers Question 11.
Jermaine paid $37.95 for 11 gallons of gasoline. What was the price in dollars per gallon?

Answer:
To find a price in dollars per gallon, divide $37.95 by 11.

Therefore, each gallon cost $3.45.

Hot Tip! Pay attention to the units given in a test question, especially if there are mixed units, such as inches and feet.

Question 12.
Shown below is a scale drawing of a rectangular patio.

Answer:
2 cm : 1 ft ⇔ 1 cm : \(\frac{1}{2}\) ft
P = perimeter of a rectangle
l = 21 cm
w = 12 cm
Use the formula for the perimeter of a rectangle.
P = 2(l + w) Substitute 21 for l and 12 for w.
P = 2(21 + 12)
P = 2.33
P = 66 cm
Because 1 cm : \(\frac{1}{2}\) ft, the perimeter of the actual patio is 66 ∙ \(\frac{1}{2}\) = 33 ft.

Texas Go Math Grade 7 Unit 2 Vocabulary Preview Answer Key

Use the puzzle to preview key vocabulary from this unit. Unscramble the circled letters within found words to answer the riddle at the bottom of the page.

1. A relationship between two quantities in which the rate of change or the ratio of one quantity to the other is constant. (Lesson 2-2)

2. Describes how much a quantity decreases in comparison to the original amount. (Lesson 3-2)

3. A fixed percent of the principal. (Lesson 3-4)

4. Angles of two or more similar shapes that are in the same relative position. (Lesson 4-1)

5. A proportional two-dimensional drawing of an object. (Lesson 4-3)

Question.
What did the athlete order when he needed a huge helping of mashed potatoes?
Answer:
_____ ____ ____ ____ – ____ ____ ____ ____ ____ ____ ____!

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 4.2 Answer Key Using Similar Shapes.

Texas Go Math Grade 7 Lesson 4.2 Answer Key Using Similar Shapes

Example 1

△ABC ~ △JKL. Find the unknown measures.
A. Find the unknown side, x.

B. Find y.
∠K corresponds to ∠B
y = ∠103°
Corresponding angles of similar triangles have equal angle measures.

Reflect

Texas Go Math Grade 7 Answers Lesson 4.2 Question 1.
Analyze Relationships What other proportion could be used to find the value of x in the example? Explain.
Answer:
We could use the proportion between the other two sides that were not used: AC and JL and do the same calculation.
\(\frac{A C}{J L}\) = \(\frac{B C}{K L}\)
\(\frac{16}{56}\) = \(\frac{12}{x}\)
\(\frac{16 \div 8}{56 \div 8}=\frac{2}{7}=\frac{12}{x}\)
2x = 7 × 12
2x = 84
x = 42 cm

Your Turn

△ABC ~ △JGH. Find the unknown measures.

Question 2.

Answer:
∠y corresponds to ∠BCA because they are both on the right from the right angle.
∠y = ∠BCA = 59°
Write a proportion using corresponding sides.
AB corresponds to GJ, and BC corresponds to GH.
\(\frac{A B}{G J}\) = \(\frac{B C}{G H}\)
\(\frac{10}{5}\) = \(\frac{6}{x}\)
10x = 5 × 6
10x = 30
x = 3 cm

Question 3.

Answer:
∠s corresponds to ∠HJG because they are obtuse angles. A triangle can have only one obtuse angle.
∠s = ∠HJG = 120°
Write a proportion using the corresponding sides
AB corresponds to GJ, and AC corresponds to HJ.

Go Math 7th Grade Answer Key Pdf Similar Shapes Question 4.
These rectangular gardens are similar in shape. Find the width of the smaller garden.

Answer:
Write proportions between corresponding sides

Question 5.
A student who is 4 feet tall stands beside a tree. The tree has a shadow that is 12 feet long at the same time that the shadow of the student is 6 feet long. Find the height of the tree. _________
Answer:
Write a proportion

The tree is 8 ft tall.

Question 6.
A photographer is taking a photo of a statue of Paul Bunyan, the legendary giant lumberjack. He measures the length of his shadow and the shadow cast by the statue. Find the height of the

Paul Bunyan statue. ___________
Answer:
Write a proportion

The statue of Paul Bunyan is 18 ft tall.

Texas Go Math Grade 7 Lesson 4.2 Guided Practice Answer Key

△ABC ~ △XYZ in each pair. Find the unknown measures (Example 1)

Question 1.

Answer:
∠d corresponds to ∠XZY because they are both on the right from the right angLe
∠d = ZXZY = 53°
Write a proportion using the corresponding sides.
AC corresponds to XZ, and AB corresponds to XY
\(\frac{A C}{X Z}\) = \(\frac{A B}{X Y}\)
\(\frac{6}{15}\) = \(\frac{8}{z}\)
6z = 15 × 8
6z = 120
z = 20 m

Go Math 7th Grade Pdf Angles of Triangles Answer Key Question 2.

Answer:
∠s corresponds to ∠CBA.
∠s = ∠CBA = 85°
Write a proportion using the corresponding sides
AB corresponds to XY, and BC corresponds to YZ.

Question 3.

Answer:
∠x corresponds to ∠ACB.
∠x = ∠ACB = 46°
Write a proportion using corresponding sides.
AC corresponds to XZ, and AB corresponds to XY.
\(\frac{X Z}{A C}\) = \(\frac{X Y}{A B}\)
\(\frac{44}{11}\) = \(\frac{t}{8}\)
4 = \(\frac{t}{8}\)
t = 4 × 8
t = 32 in.

Question 4.

Answer:
∠a corresponds to ∠YXZ.
∠a = ∠YXZ = 84°
Write a proportion using the corresponding sides
AC corresponds to XZ, and BC corresponds to YZ.
\(\frac{X Z}{A C}\) = \(\frac{Y Z}{B C}\)
\(\frac{72}{18}\) = \(\frac{x}{25}\)
x = \(\frac{x}{25}\)
x = 4 × 25
x = 100 ft

The rectangles in each pair are similar. Find the unknown measures. (Example 1)

Question 5.

Answer:
Al angles in a rectangle (in this case square) are equal to 90°.
Thus, ∠x = 90°.
Since we concluded the second rectangle is a square (all sides are equal), then the first rectangle is a square too.
Thus, y = 9.5 in

Go Math Grade 7 Lesson 4.2 Answer Key Question 6.

Answer:
All angles in a rectangle are equal to 90°.
Thus, ∠f = 90°.
Write a proportion using the corresponding sides.
RS corresponds to BC, and QR corresponds to AB.

q = 10 cm

Question 7.
Samantha wants to find the height of a pine tree in her yard. She measures the height of the mailbox at 3 feet and its shadow at 4.8 feet. Then she measures the shadow of the tree at 56 feet. How tall is the tree? (Example 3)
Answer:
Write a proportion

4.8h = 3 × 56
4.8h = 168
h = 35 feet
The height of the pine tree is 35 feet.

Essential Question Check-In

Question 8.
Describe how you could use your height and a yardstick to determine the unknown height of a flagpole.
Answer:
We could measure your height, your shadow and the flagpole’s shadow.
Then, we would write a proportion using thoes measures.

Texas Go Math Grade 7 Lesson 4.2 Independent Practice Answer Key

Question 9.
A cactus casts a shadow that is 15 ft long. A gate nearby casts a shadow that is 5 ft long. Find the height of the cactus.

Answer:
Write a proportion

The height of the cactus is 9 ft.

Question 10.
Two ramps modeled with triangles are similar. Which side of triangle KOA corresponds to \(\overline{J N}\)? Explain.

Answer:
We can see that \(\overline{J N}\) Lies on the left side of the right angLe. In triangle KOA \(\overline{K O}\) Lies on the Left side of the right angle.
Thus, \(\overline{K O}\) responds to \(\overline{J N}\).

Question 11.
A building with a height of 14 m casts a shadow that is 16 m long while a taller building casts a 24 m long shadow. What is the height of the taller building?
Answer:
Write a proportion

The taller building is 24 m tall.

Texas Go Math Grade 7 Similar Triangles Pdf Question 12.
Katie uses a copy machine to enlarge her rectangular design that is 6 in. wide and 8 in. long. The new width is 10 in. What is the new length?
Answer:
Write a proportion

The new length is \(13 . \overline{3}\) in.

Question 13.
Art An art exhibit at a local museum features several similarly shaped metal cubes welded together to make a sculpture. The smallest cube has a edge length of 6 inches.
a. What are the edge lengths of the other cubes if the ratios of similarity to the smallest cube are 1.25, \(\frac{4}{3}\), 1.5, \(\frac{7}{4}\), and 2 respectively?
Answer:
l₁ = length of the smallest cube = 6 in.
Calculate lengths of other cubes using given ratios.

b. If the artist wanted to add a smaller cube with an edge length with a ratio of \(\frac{2}{3}\) to the sculpture, what size would the cube be?
Answer:
l_a = added cube
Again, use the given ratio to calculate l_a
\(\frac{l_{a}}{l_{1}}\) = \(\frac{2}{3}\)
\(\frac{l_{a}}{6}\) = \(\frac{2}{3}\)
l_a = \(\frac{2}{3}\) × 6
l_a = 4 in.

c. Why do you only have to find the length of one edge for each cube?
Answer:
We only have to find length of one edge for each cube, because all edges of a cube are of equal length.

△XYZ ~ △PQR in each pair. Find the unknown measures.

Question 14.

Answer:
∠b corresponds to ∠XZY
∠d = ∠XZY = 89°
Write a proportion using the corresponding sides.
XZ corresponds to PR, and YZ corresponds to RQ.
\(\frac{P R}{X Z}\) = \(\frac{R Q}{Y Z}\)
\(\frac{20}{8}\) = \(\frac{a}{9}\)
\(\frac{5}{2}\) = \(\frac{a}{9}\)
2a = 5 × 9
2a = 45
a = 22.5 cm

Question 15.

Answer:
∠s corresponds to ∠RQP.
∠s = ∠RQP = 58°
Write a proportion using the corresponding sides.
x corresponds to and XZ corresponds to PR.

Question 16.
Two common envelope sizes are 3\(\frac{1}{2}\) in. × 6\(\frac{1}{2}\) in. and 4in. × 9\(\frac{1}{2}\) in. Are these envelopes similar? Explain.
Answer:
Check if the ratios of the corresponding sides are equal.

Although all corresponding angles of enveLopes are equal, the ratios of corresponding sides are not equal.
Thus, the envelopes are not similar

Go Math 7th Grade Lesson 4.2 Find the Value of the Shapes Question 17.
A pair of rectangular baking pans come in a set together for $15. One pan is 13 inches by 9 inches and the other pan is 6 inches by 6 inches. Without doing any calculations, how can you tell that these pans are not similar?
Answer:
We can see right away that the sides of the second pan are equal. Thus, the second pan has the shape of a square white the first one has the shape of a rectangle.
From this, we can conclude that the pans are not similar.

H.O.T. Focus On Higher Order Thinking

Question 18.
Draw Conclusions In the similar triangles used in indirect measurement with the shadows of a flagpole and a person, which sides of the triangles represent the rays of the sun?
Answer:
The hypotenuse represents the rays of the sun.

Question 19.
Make a Conjecture Do you think it is possible to use indirect measurement with shadows if the sun is directly overhead? Explain.
Answer:
No, it is not possible.
When the sun is directly overhead, the Length of the shadow is 0. Thus, comparing any ratio including shadow side in it would equal to 0 (\(\frac{0}{x}\)), or would not be possible to divide (\(\frac{x}{0}\))

Question 20.
Analyze Relationships Joseph’s parents have planted two gardens. One is square and has an area of 25 ft². The other one has two sides equal to \(\frac{2}{3}\) of one side of the square, and the other two sides equal to \(\frac{5}{2}\) of one side of the square.

a. Find the dimensions of the other garden. Explain how you found your answer.
Answer:
If first garden ¡s a square, that means his area is calculated with the following formula:
A = a²
Now, calculate its side using that A = 25 ft².
25 = a²
a = \(\sqrt {25}\)
a = 5 ft
Next, calculate the sides of the other garden using given ratios.
x = first and third side of the other garden
y = second and fourth side of the other garden
x = \(\frac{2}{3}\) × 5
x = \(\frac{10}{3}\) ft

y = \(\frac{5}{2}\) × 5
y = \(\frac{25}{2}\) ft

b. Find the area of the other garden.
Answer:
Calculate the area of the other garden, which is a rectangle, with the following formula:
A = x × y
A = \(\frac{10}{3}\) × \(\frac{25}{2}\)
A = \(\frac{5 \times 25}{3 \times 1}\)
A = \(\frac{125}{3}\) ft²

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 7 Answer Key Linear Relationships.

Texas Go Math Grade 7 Module 7 Answer Key Linear Relationships

Texas Go Math Grade 7 Module 7 Are You Ready? Answer Key

Evaluate each expression.

Question 1.
2(5) + 11 __________
Answer:
2(5) + 11 = 10 + 11 Multiply 2 by 5.
= 21 Add 11 to this product.
2(5) + 11 = 21

Go Math Grade 7 Module 7 Answer Key Question 2.
9(6) – 5 __________
Answer:
9 (6) – 5 = 54 – 5 Multiply 9 by 6.
= 49 Subtract 5 from this product
9(6) – 5 = 49

Question 3.
4(12) – 15 __________
Answer:
4 (12) – 15 = 48 – 15 Multiply 4 by 12.
= 33 Subtract 15 from this product.
4(12) – 15 = 33

Question 4.
-6(2) + 13 ___________
Answer:
-6(2) + 13 = -12 + 13 Multiply -6 by 2.
= 1 Add 13 to this product.
-6(2) + 13 = 1

Question 5.
7(-4) – 8 ____________
Answer:
7(- 4) – 8 = -28 – 8 Multiply 7 by -4.
= -36 Since both numbers are negative, they are added together with the -“ sign.
7(-4) – 8 = -36

Question 6.
-2(-5) + 7 ____________
Answer:
-2(- 5) + 7 = 10 + 7 Multiply -2 by -5.
= 17 Add 7 to this product.
-2(-5) + 7 = 17

Find a rule relating the given values.

Question 7.

Answer:
Set the equation for x = 1, y = 3. For all the others apply the same rule.
1 ∙ z = 3
Notice that this variable z is 3 because;
1 ∙ 3 = 3
Therefore the rule for the given value is:
y = x ∙ 3

Grade 7 Module 7 Inequalities Answer Key Question 8.

Answer:
Set the equation for x = 1, y = 9. For all the others apply the same rule
1 + z = 9
Subtract one from each side.
z = 1 – 1 + z = 9 – 1 = 8
Therefore the rule for the given value is:
y = x + 8

Graph point A(4, 3). Start at the origin. Move 4 units right. Then move 3 units up.

Graph each point on the coordinate grid above.

Question 9.
B(9, 0)
Answer:
For graphing a point A = (9, 0) starting from the origin move 9 units right

Linear Relationships Homework 7 Answer Key Question 10.
C(2, 7)
Answer:
C = (2, 7)
Starting from the origin move 2 units right and then move 7 units up.

Question 11.
D(0, 5)
Answer:
D = (0, 5)
Starting from the origin move 5 units up.

Module 7 Answer Key Solving Linear Equations Question 12.
E(6, 2)
Answer:
E = (6, 2)
Starting from the origin move 6 units right and then move 2 units up.

Texas Go Math Grade 7 Module 7 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the third column of the chart.

Understand Vocabulary

Answer each question with the correct preview word.

Question 1.
What is a mathematical statement that two expressions are equal?
Answer:
An equation is a mathematical, statement that two expressions are equal.

Grade 7 Go Math Module 7 Answer Key Question 2.
What is a constant ratio of two variables that are related proportionally?
Answer:
The constant of proportionally is a constant ratio of two variables related proportionally.

Active Reading
Tri-Fold Before beginning the module, create a tri-fold to help you learn the concepts and vocabulary in this module. Fold the paper into three sections. Label the columns” What I Know:’ “What I Need to Know”, and “What I Learned.” Complete the first two columns before you read. After studying the module, complete the third column.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 4.3 Answer Key Similar Shapes and Scale Drawings.

Texas Go Math Grade 7 Lesson 4.3 Answer Key Similar Shapes and Scale Drawings

Texas Go Math Grade 7 Lesson 4.3 Explore Activity Answer Key

Explore Activity 1

Finding Dimensions

Scale drawings and scale models are used in mapmaking, construction, and other trades.

A blueprint is a technical drawing that usually displays architectural plans. Pete’s blueprint shows a layout of a house. Every 4 inches in the blueprint represents 3 feet of the actual house. One of the walls in the blueprint is 24 inches long. What is the actual length of the wall?

A. Complete the table to find the actual length of the wall.

Reflect

Question 1.
In Pete’s blueprint, the length of a side wall is 16 inches. Find the actual length of the wall.
Answer:
First, divide by 4 and then multiply by 3 to find the actual length of the wall.
16 ÷ 4 × 3 = 4 × 3
= 12 ft
The actual length of the wall is 12 ft

Similar Shapes and Scale Drawings Worksheet 7th Grade Pdf Question 2.
The back wall of the house is 33 feet long. What is the length of the back wall in the blueprint?
Answer:
Now, do the opposite. First, divide by 3 and then multiply by 4 to find the length of the wall in the blueprint.
33 ÷ 3 × 4 = 11 × 4
= 44 in.
The length of the wall in the blueprint is 44 in.

Question 3.
Check for Reasonableness How do you know your answer to 2 is reasonable?
Answer:
I know my answer to 2 is reasonable because when I revert the process, I get that the actual length is 33 ft.
44 ÷ 4 × 3 = 11 × 3
= 33 ft.

Example 1

The art class is planning to paint a mural on an outside wall. This figure is a scale drawing of the wall. What is the area of the actual wall?

STEP 1: Find the number of feet represented by 1 inch in the drawing.
\(\frac{2 \mathrm{in} \cdot \div 2}{3 \mathrm{ft} \div 2}=\frac{1 \mathrm{in}}{1.5 \mathrm{ft}}\)
1 inch in this drawing equals 1.5 feet on the actual wall.

STEP 2: Find the height of the actual wall labeled 11 inches in the drawing.
\(\frac{1 \text { in. } \times 11}{1.5 \mathrm{ft} \times 11}=\frac{11 \mathrm{in} .}{16.5 \mathrm{ft}}\)
The height of the actual wall is 16.5 ft.

STEP 3: Find the length of the actual wall labeled 28 inches in the drawing.
\(\frac{1 \mathrm{in} \cdot \times 28}{1.5 \mathrm{ft} \times 28}=\frac{28 \mathrm{in} .}{42 \mathrm{ft}}\)
The length of the actual wall is 42 ft.

STEP 4: Since area is length times width, the area of the actual wall is 16.5 ft × 42 ft = 693 ft².

Reflect

Go Math Grade 7 Lesson 4.3 Answer Key Question 4.
Analyze Relationships. Flow could you solve the example without having to determine the number of feet represented by 1 inch?
Answer:
We could simply multiply the height and length by \(\frac{2 \text { in. }}{3 \mathrm{ft} .}\).

Your Turn

Question 5.
Find the length and width of the actual room, shown in the scale drawing. Then find the area of the actual room. Round your answer to the nearest tenth.

Answer:
Multiply length and width by the scale.
l = 6.5 in. × \(\frac{8 \mathrm{~ft}}{3 \text { in. }}\)
= \(\frac{52}{3}\)
= \(17 . \overline{3}\) ft

w = 5 in. × \(\frac{8 \mathrm{~ft}}{3 \text { in. }}\)
= \(\frac{40}{3}\)
= \(13 . \overline{3}\) ft

Now. calculate the area using the following formula: A = l × w
A = l × w
= \(17 . \overline{3}\) × \(13 . \overline{3}\)
= \(231 . \overline{1}\) = (round to the nearest tenth) = 231.1 ft²

Question 6.
The drawing plan for an art studio shows a rectangle that is 13.2 inches by 6 inches. The scale in the plan is 3 in.:5 ft. Find the length and width of the actual studio. Then find the area of the actual studio.
Answer:
Multiply length and width by the scale.
l = 13.2 in. × \(\frac{5 \mathrm{~ft}}{3 \mathrm{~in}}\)
= 22 ft

w = 6 in. × \(\frac{5 \mathrm{ft}}{3 \mathrm{in}}\)
= 10 ft

Now. calculate the area using the following formula: A = l × w
A = l × w
= 22 × 10
= 220 ft²

Explore Activity 2

Drawing in Different Scales
A. A scale drawing of a meeting hail is drawn on centimeter grid paper as shown. The scale is 1 cm:3 m.
Suppose you redraw the rectangle on centimeter grid paper using a scale of 1 cm:6 m. In the new scale, 1 cm represents more than/less than 1 cm in the old scale.
The measurement of each side of the new drawing will be twice/half as long as the measurement of the original drawing.

B. Draw the rectangle for the new scale 1 cm:6 m.

Reflect

Go Math Lesson 4.3 7th Grade Similar Shapes and Scale Drawings Question 7.
Find the actual length of each side of the hall using the original drawing. Then find the actual length of each side of the hall using the your new drawing and the new scale. Flow do you know your answers are correct?
Answer:
First measure length and width on the scale drawing.
l = 4cm
w = 3 cm
Now, find the actual length and width using the scale.
l = 4 × 3 = 12 m
w = 3 × 3 = 9 m
Next find the actual length and width using the scale and new drawing
l = 2 × 6 = 12 m
w = 1.5 × 6 = 9m
I know my answers are correct because I get the same actual length and width using both drawings.

Texas Go Math Grade 7 Lesson 4.3 Guided Practice Answer Key

Question 1.
The scale of a room in a blueprint is 3 in : 5 ft. A wall in the same blueprint is 18 in. Complete the table. (Explore Activity 1)

Answer:

Complete the table by multiplying all lengths by the scale.


a. How long is the actual wall?
Answer:
Check the table. The actual wall is 30 ft long.

b. A window in the room has an actual width of 2.5 feet. Find the width of the window in the blueprint.
Answer:
Multiply the width by the scale.
w = 2.5 ft × \(\frac{3 \text { in. }}{5 \mathrm{~ft}}\)
= 1.5 in.
The actual width of the window is 1.5 in.

Question 2.
The scale in the drawing is 2 in.: 4 ft. What are the length and 14 in. width of the actual room? Find the area of the actual room. (Example 1)

Answer:
Multiply length and width by the scale.
l = 14 in. × \(\frac{4 \mathrm{~ft}}{2 \mathrm{~in} .}\)
= 28 ft

w = 7 in. × \(\frac{4 \mathrm{~ft}}{2 \mathrm{~in} .}\)
= 11 ft
Now. calculate the area using the following formula: A = l × w
A = l × w
= 28 × 11
= 392 ft²

Go Math Lesson 4.3 Answer Key 7th Grade Question 3.
The scale in the drawing is 2 cm : 5 m. What are the length and width 10 cm of the actual room? Find the area of the actual room. (Example 1)

Answer:
Multiply length and width by the scale.
l = 10 cm × \(\frac{5 \mathrm{~m}}{2 \mathrm{~cm}}\)
= 25 m

w = 6 cm × \(\frac{5 \mathrm{~m}}{2 \mathrm{~cm}}\)
= 15 m
Now, calculate the area using the following formula: A = l × w
A = l × w
= 25 × 15
= 375 ft²

Question 4.
A scale drawing of a cafeteria is drawn on centimeter grid paper as shown. The scale is 1 cm : 4 m. (Explore Activity 2)
a. Redraw the rectangle on centimeter grid paper using a scale of 1 cm:6 m.

Answer:
First, find length and width on the scale drawing.
l = 4.5 cm
w = 3 cm.
Now, find the actual dimensions and then use scale to find dimensions to redraw the rectangle.
Actual length using original scale length: l = 4.5 cm × \(\frac{4 \mathrm{~m}}{1 \mathrm{~cm}}\)
= 18 m
New scale length using actual length: l = 18m × \(\frac{1 \mathrm{~cm}}{6 \mathrm{~m}}\)
= 3 cm

Actual width using original scale width: w = 3 cm × \(\frac{4 \mathrm{~m}}{1 \mathrm{~cm}}\)
= 12 m
New scale width using actual width: w = 12 m × \(\frac{1 \mathrm{~cm}}{6 \mathrm{~m}}\)
= 2 m
Redraw the rectangle using new scale length and width. (Picture below)

b. What is the actual length and width of the cafeteria using the original scale? What are the actual dimensions of the cafeteria using the new scale?
Answer:
The actual length and width using original scale and new scale are the same.
l = 18m
w = 12w

Essential Question Check-In

Scale Drawings Worksheet 7th Grade Answers Question 5.
If you have an accurate, complete scale drawing and the scale, which measurements of the object of the drawing can you find?
Answer:
We can find actual measurements of the object of the drawing.

Texas Go Math Grade 7 Lesson 4.3 Independent Practice Answer Key

Question 6.
Art Marie has a small copy of Rene Magritte’s famous painting, The Schoolmaster. Her copy has dimensions 2 inches by 1.5 inches. The scale of the copy is 1 in:40 cm.
a. Find the dimensions of the original painting.
Answer:
Multiply the dimensions by the scale to get the dimensions of the original painting.
l = 2 in. × \(\frac{40 \mathrm{~cm}}{1 \mathrm{~in} .}\)
= 2 in. × \(\frac{40 \mathrm{~cm}}{1 \mathrm{~in} .}\)
= 80 cm

w = 1.5 in. × \(\frac{40 \mathrm{~cm}}{1 \mathrm{~in} .}\)
= 1.5 in. × \(\frac{40 \mathrm{~cm}}{1 \mathrm{~in} .}\)
= 60 cm

b. Find the area of the original painting.
Answer:
Calculate the area using the following equation: A = l × w
A = l × w
= 80 × 60
= 4800 cm²

c. Since 1-inch is 2.54 centimeters, find the dimensions of the original painting in inches.
Answer:
Find the dimensions of the original painting in niches by multiplying the dimensions by the conversion factor.
Conversion factor: \(\frac{1 \mathrm{~in} .}{2.54 \mathrm{~cm}}\)
l = 80 cm × \(\frac{1 \mathrm{~in} .}{2.54 \mathrm{~cm}}\)
= 80 cm × \(\frac{1 \mathrm{~in} .}{2.54 \mathrm{~cm}}\)
≈ 31.5 in.

w = 60 cm × \(\frac{1 \mathrm{~in} .}{2.54 \mathrm{~cm}}\)
= 60 cm × \(\frac{1 \mathrm{~in} .}{2.54 \mathrm{~cm}}\)
≈ 23.6 in.

d. Find the area of the original painting in square inches.
Answer:
Calculate the are using the following equation: A = l × w
A = l × w
≈ 31.5 × 23.6
≈ 743.4 in.²

Question 7.
A game room has a floor that is 120 feet by 75 feet. A scale drawing of the floor on grid paper uses a scale of 1 unit:5 feet. What are the dimensions of the scale drawing?
Answer:
Multiply actual dimensions by the scale to get the dimensions of the scale drawing.

Lesson 4.3 Similar Shapes and Scale Drawings Answer Key Question 8.
Multiple Representations The length of a table is 6 feet. On a scale drawing, the length is 2 inches. Write three possible scales for the drawing.
Answer:
Divide the length on a scale drawing by the actuaL Length to get the scale.
2 inches : 6 feet
Now, we can write it as a fraction, and expand or simplify to get 2 more possible scales.

We got 2 more possible scales:
1 inches : 3 feet
4 inches : 12 feet

Question 9.
Analyze Relationships A scale for a scale drawing is 10 cm:1 mm. Which is larger, the actual object or the scale drawing? Explain.
Answer:
The scale drawing is larger, because 10 cm on the scale drawing is 1 mm on the actual object

Question 10.
Architecture The scale model of a building is 5.4 feet tall.
a. If the original building is 810 meters tall, what was the scale used to make the model?
Answer:
Divide the scale modeL height by the actual height to get the scale used to make the model.
\(\frac{5.4 \text { feet }}{810 \text { meters }}=\frac{5.4 \div 5.4 \text { feet }}{810 \div 5.4 \text { meters }}\)
= \(\frac{1 \text { foot }}{150 \text { meters }}\)
The scale used to make the model was: 1 foot : 150 meters

b. If the model is made out of tiny bricks each measuring 0.4 inch in height, how many bricks tall is the model?
Answer:
Divide the scale model height by the brick height to get how many bricks talL is the model.
5.4 ÷ 0.4 = 13.5
The scale model is 13.5 bricks tall.

Question 11.
You have been asked to build a scale model of your school out of toothpicks. Imagine your school is 30 feet tall. Your scale is 1 ft:1.26 cm.
a. If a toothpick is 6.3 cm tall, how many toothpicks tall will your model be?
Answer:
First, find out how many cm is your scale model tall by multiplying actual height by the scale.
h = 30 ft × \(\frac{1.26 \mathrm{~cm}}{1 \mathrm{~ft}}\)
= 30 ft × \(\frac{1.26 \mathrm{~cm}}{1 \mathrm{~ft}}\)
= 37.8 cm
Now, divide obtained scale model height by toothpick height to get how many toothpick tall our scale model will be.
37.8 ÷ 6.3 = 6
Our scale model will be 6 toothpicks tall.

b. Your mother is out of toothpicks, and suggests you use cotton swabs instead. You measure them, and they are 7.6 cm tall. How many cotton swabs tall will your model be?
Answer:
Divide the scale model height obtained in the subtask above by cotton swabs height to get how many cotton swabs tall our scale model will be.
37.8 ÷ 7.6 ≈ 5
Our scale model will be ≈ 5 cotton swabs tall.

H.O.T. Focus On Higher Order Thinking

Question 12.
Draw Conclusions The area of a square floor on a scale drawing is 100 square centimeters, and the scale of the drawing is 1 centimeter:2 ft, What is the area of the actual floor? What is the ratio of the area in the drawing to the actual area?
Answer:
First, calculate the length of the floor on a scale drawing using the fact it is a square floor.
A = l²
100 = l²
l = \(\sqrt {100}\)
l = 10 cm

Now, calculate the actual length of the square floor by multiplying the scale drawing length by the scale.
l = 10 cm × \(\frac{2 \mathrm{~ft}}{1 \mathrm{~cm}}\)
= 10 cm × \(\frac{2 \mathrm{~ft}}{1 \mathrm{~cm}}\)
= 20 ft
Next, calculate the actual area of the floor.
A = l²
= 20²
= 400 ft²
Write the ratio of the areas.
\(\frac{100 \mathrm{~cm}^{2}}{400 \mathrm{ft}^{2}}=\frac{1 \mathrm{~cm}^{2}}{4 \mathrm{ft}^{2}}\)
The ratio of the area in the drawing to the actual area is 1 : 4.

Go Math 7th Grade Scale Drawing Assessment Answer Key Question 13.
Multiple Representations Describe how to redraw a scale drawing with a new scale.
Answer:
First, calculate the actual dimensions by multiplying originaL scale drawing dimensions by the original scale. Then, calculate new scale drawing dimensions by multiplying actual dimensions by the new scale.

Question 14.
Represent Real-World Problems Describe how several jobs or professions might use scale drawings at work.
Answer:
Architects use scale drawings to draw a model of the house they are designing.
Engineers use scale drawings to draw an engine they are building.
Fashion designers use scale drawings to draw an outfit before it is made.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 4.1 Answer Key Similar Shapes and Proportions.

Texas Go Math Grade 7 Lesson 4.1 Answer Key Similar Shapes and Proportions

Texas Go Math Grade 7 Lesson 4.1 Explore Activity Answer Key

Similar Shapes and Proportions

Similar shapes have the same shape but not necessarily the same size. You can use square tiles to model similar figures.

A rectangle made of square tiles measures 5 tiles long and 2 tiles wide. Find the length of a similar rectangle that measures 6 tiles wide.

STEP 1: Using the square tiles, make a rectangle 5 tiles long and 2 tiles wide.

STEP 2: Add tiles to increase the width of the rectangle to 6 tiles.
There are now _____ sets of the original _____ tiles along the width of the rectangle because _____ × _____ = 6.
The width of the rectangle is _____ times the width of the original rectangle.

STEP 3: Add tiles to also increase the length of the rectangle.
The width of the new rectangle is _____ times the width of the original. To keep the lengths of the sides
proportional, the length must also be _____ times the length of the original. The length of the similar rectangle is _____ × ____ = 15.
The length of the similar rectangle is ________ tiles.

Reflect

Similar Figures and Proportions Lesson 4.1 Answer Key Geometry Question 1.
Justify Reasoning If one dimension is changed, why does the other dimension have to change to create a similar figure?
Answer:
The other dimension has to change because the ratio between dimensions of a similar figure has to remain the same

Your Turn

Explain whether the triangles are similar.

Question 2.

Answer:
The corresponding angles of a triangle do not have equal measures. Thus, the triangles are not similar.

Question 3.

Answer:
Check that the corresponding angles of the triangle have equal measures.
m∠A = m∠D = 33°
m∠B = m∠E = 129°
m∠C = m∠F = 18°
Corresponding angles ensure that the corresponding sides will be proportional Thus, the triangles are similar

Your Turn

Explain whether the shapes are similar.

Lesson 4.1 Similar Figures Answer Key Go Math Grade 7 Question 4.
rectangle ABCD with sides of 7 and 5 and rectangle MNOP with sides of 21 and 15
Answer:
Let \(\overline{A B}\) = 7, \(\overline{C D}\) = 7, \(\overline{B C}\) = 5, \(\overline{A D}\) = 5
Let \(\overline{M N}\) = 21, \(\overline{O P}\) = 21, \(\overline{N O}\) = 15, \(\overline{M P}\) = 15.
Angles are corresponding ALL are equal to 90°.
\(\overline{A B}\), \(\overline{C D}\) correspond to \(\overline{M N}\), \(\overline{O P}\) respectively.
\(\overline{B C}\), \(\overline{A D}\) correspond to \(\overline{N O}\), \(\overline{M P}\) respectively.

Since the measures of the corresponding angles are equal and the corresponding sides are proportional, the rectangles are proportional.

Question 5.

Answer:
It is obvious from the scale drawing that the corresponding angles are equal.
Now, let’s check for the sides

The shapes are not similar because the ratios between corresponding sides are not equal.

Texas Go Math Grade 7 Lesson 4.1 Guided Practice Answer Key

Question 1.
A rectangle made of square tiles measures 7 tiles long and 3 tiles wide. What is the length of a similar rectangle whose width is 9 tiles? (Explore Activity)
Answer:
Ratios between dimensions have to be preserved.
\(\frac{7}{l}\) = \(\frac{3}{9}\)
7 × 9 = 3l
3l = 63
l = 21
Length of a similar rectangle whose width is 9 tiles is 21 tiles.

Explain whether the shapes are similar. (Examples 1 and 2)

Question 2.

Answer:
We can see without looking whether the sides correspond that the angles do not correspond. Thus, the triangles are not similar.

Lesson 4.1 Geometry Answers Go Math Grade 7 Question 3.

Answer:
The angles are corresponding.
We can see that both shapes have the same length of sides. We can conclude that both shapes are squares.
Thus, the shapes are similar.

Essential Question Check-In

Question 4.
Describe how to use ratios to determine whether two shapes are similar.
Answer:
First, check if the angles are corresponding. Then, check if the ratios of the corresponding sides are equal.

Texas Go Math Grade 7 Lesson 4.1 Independent Practice Answer Key

Determine if each statement is true or false. Justify your answer.

Question 5.
All squares are similar. __________
Answer:
Let’s check what it takes for a quadrangle to be a square.

1. All of the quadrangle’s sides must have the same length.
2. All of the quadrangle’s angles must be equal to 90°

Now, it is obvious that the angles are corresponding. Next, we can see that the ratios of corresponding sides are obviously equal because all 4 sides of a square are of equal length. Thus, all squares are similar.

Question 6.
All right triangles are similar. __________
Answer:
False

A triangle needs to have one right angle to be a right triangle That does not mean the other 2 angles are corresponding. Thus, all right angles are not similar.

Art For 7-10, use the table. Assume all angle measures are equal to 90°.

Go Math Grade 7 Lesson 4.1 Answer Key Question 7.
Hugo has a small print of one of the paintings on the table. It is similar in size to the original. The print measures 11 in. × 10 in. Of which painting is this a print? Explain.
Answer:
It is a print of “The Dance Class” because the dimensions are corresponding
\(\frac{11}{33}\) = \(\frac{10}{30}\)
\(\frac{1}{3}\) = \(\frac{1}{3}\)

Question 8.
A local artist painted a copy of Cezanne’s painting. It measures 88 in. × 74 in. Is the copy similar to the original? Explain.
Answer:
Check if the ratios between corresponding dimensions are equal

The copy is not similar to the original.

Question 9.
A company made a poster of da Vinci’s painting. The poster is 5 feet 3.5 feet wide. Is the poster similar to the original Mona Lisa? Explain.
Answer:
Check if the ratios between corresponding dimensions are equal

The copy is similar to the original.

Lesson 4.1 Similar Shapes and Proportions Answer Key Question 10.
The same company made a poster of The Blue Vase. The poster is 36 inches long and 26 inches wide. Is the poster similar to the original The Blue Vase? Explain.
Answer:
Check if the ratios between corresponding sides are equal

The copy is not similar to the original.

Problem Solving The figure shows a 12 ft by 15 ft garden divided into four rectangular parts, each planted with a different vegetable. Explain whether the rectangles in each pair are similar and why.

Question 11.
rectangle A and the original rectangle
Answer:
Check if the ratios between corresponding sides are equal
Original rectangLe: 12 ft. by 15 ft.
Rectangle A: 4 ft. by 5 ft.
\(\frac{12}{4}\) \(\stackrel{?}{=}\) \(\frac{15}{5}\)
3 = 3
Rectangle A and the original rectangle are similar.

Question 12.
the original rectangle and rectangle D
Answer:
Check if the ratios between the corresponding sides are equal.
Original rectangle: 12 ft. by 15 ft.
Rectangle D: (12 – 4) ft. by (15 – 5) ft.
Rectangle D: 8 ft. by 10 ft.

Rectangle D and the original rectangle are similar.

Question 13.
rectangle C and rectangle B
Answer:
Check if the ratios between the corresponding sides are equal.
Rectangle B: 5 ft. by (12 – 4) ft.
Rectangle B: 5 ft. by 8 ft.
Rectangle C: 4 ft. by (15 – 5) ft.
Rectangle C: 4 ft. by 10 ft.
\(\frac{5}{4}\) ≠ \(\frac{8}{10}\)
Observe that the fraction on the left side is greater than 1, and the fraction on the right side is lesser that 1. We can conclude that they are not equal without further calculation.
Rectangle B and rectangle C are not similar.

H.O.T. FOCUS ON HIGHER ORDER THINKING

Question 14.
Analyze Relationships Which of these four-sided shapes are similar?

Answer:
First, compare the first and the second shape.
They both have all 4 angles equal to 90°
Now, we can check the ratios between corresponding sides.

The first and the second shape are similar
Now, we can see that the first and the third shape have equal corresponding sides. Thus, the ratios between corresponding sides are equal.

But, are they similar? The answer is no Because the corresponding angles are obviously not equal.

The first and the third shape are not similar.

Now, we can just look at the angles in the second and the third shape. We can see that the corresponding angles are not equal.

The second and the third shape are not similar.

Similar Figures Proportions Go Math Grade 7 Lesson 4.1 Question 15.
Communicate Mathematical Ideas Describe the two tests two polygons must pass to be proven similar.
Answer:
First test: Two polygons must have equal corresponding angles.
Second test: Two polygons must have equal ratios between their corresponding sides.

Question 16.
Make a Conjecture Using what you know of similar figures, explain whether you believe all rectangles are similar. Give an example or a counter-example.
Answer:
All rectangles have equal corresponding angles.
But, they do not necessarily have to have equal ratios between their corresponding angles.
Check the ratios between corresponding sides from the rectangles on the picture.
\(\frac{5}{4}\) ≠ \(\frac{2}{3}\)
We can see right away that \(\frac{5}{4}\) > 1 and \(\frac{2}{3}\) < 1. Thus, they are not equal.
We have found a counter-example.