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Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 21.4 Symmetry and Transformations to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 21.4 Symmetry and Transformations

Exercises

IDENTIFY

For each pair of figures, label as translation, rotation, reflection, or dilation.

Question 1.

Answer:
Translation,

Explanation:
Both the figures are translated of one other because one is a translation of other that is identical duplicate of the other.

Question 2.

Answer:
Rotation,

Explanation:
Fiqures are rotation as a rotation is a type of transformation which is a turn. A figure can be turned clockwise or counterclockwise on the coordinate plane. In both transformations the size and shape of the figure stays exactly the same. A rotation is a transformation that turns the figure in either a clockwise or counterclockwise direction.

Question 3.

Answer:
Reflection,

Explanation:
Fiqures are reflection as a reflection is a mirror image of the shape. An image will reflect through a line, known as the line of reflection. A figure is said to reflect the other figure, and then every point in a figure is equidistant from each corresponding point in another figure.

Question 4.

Answer:
Dilation,

Explanation:
Given fiqures are dilation as a dilation is a transformation that enlarges or reduces a figure in size. This means that the preimage and image are similar and are either reduced or enlarged using a scale factor.

Question 5.

Answer:
Dilation,

Explanation:
Given fiqures are dilation as a dilation is a transformation that enlarges or reduces a figure in size. This means that the preimage and image are similar and are either reduced or enlarged using a scale factor.

Question 6.

Answer:
Rotation,

Explanation:
Fiqures are rotation as a rotation is a type of transformation which is a turn. A figure can be turned clockwise or counterclockwise on the coordinate plane. In both transformations the size and shape of the figure stays exactly the same. A rotation is a transformation that turns the figure in either a clockwise or counterclockwise direction.

Question 7.

Answer:
Reflection,

Explanation:
Fiqures are reflection as a reflection is a mirror image of the shape. An image will reflect through a line, known as the line of reflection. A figure is said to reflect the other figure, and then every point in a figure is equidistant from each corresponding point in another figure.

Question 8.

Answer:
Translation,

Explanation:
Both the figures are translated of one other because one is a translation of other that is identical duplicate of the other.

Identify lines of symmetry.

Question 9.

How many lines of symmetry does the figure above have?
Answer:
4 or four,

Explanation:
A line of symmetry is the line that divides the shape into two halves that match exactly. A square has four lines of linear symmetry. One crosses through each diagonal. The other is the two lines that cross horizontally and vertically through the middle of the square therefore a square has 4 lines of symmetry.

Question 10.

Draw the lines of symmetry on the figure above.
Answer:
6 or six,

Explanation:
A line of symmetry is the line that divides the shape into two halves that match exactly. A Hexagon has six lines of linear symmetry.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 21.3 Circles to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 21.3 Circles

Exercises

SOLVE

Question 1.
What is the diameter of the circle?

Answer:
The diameter of the circle is equal to 2 cms,

Explanation:
As we know the diameter of the circle is equal to 2 X radius, so as the given circle has
r = 1 cm so diameter = 2 x 1 cm = 2 cms.

Question 2.
What is the diameter of the circle?

Answer:
The diameter of the circle is equal to 14 in.,

Explanation:
As we know the diameter of the circle is equal to 2 X radius, so as the given circle has
r = 1 cm so diameter = 2 x 7 in. = 14 inches.

Question 3.
What is the radius of the circle below?

Answer:
The radius of the circle is equal to 4 ft,

Explanation:
As we know the diameter of the circle is equal to 2 X radius, so as the given circle has
d = 8 ft so radius = diameter/ 2 = 8 ft/2 = 4 ft.

Question 4.
What is the center of the circle?

Answer:
The center of the circle is d/2,

Explanation:
As we know the diameter of the circle is equal to 2 X radius so as the given circle has
diameter = 2 x radius or center, therefore center = d/2.

Question 5.
What is the radius of the circle?

Answer:
The radius of the circle is equal to 3 in.,

Explanation:
As we know the diameter of the circle is equal to 2 X radius, so as the given circle has
d = 6 in. so radius = diameter/ 2 = 6 in/2 = 3 in.

Question 6.
What are the two radii of the circle?

Answer:
OA,OB,

Explanation:
As a Radius or Radii of a circle is the distance from the center of the circle to any point on it’s circumference. So the given circle has two radii they are OA and OB.

SOLVE

Question 1.

What is the area of the circle? ___________
What is the circumference? ___________
Answer:
The area of the circle is 314.285 cm²,
The circumference of the circle is 62.857 cm,

Explanation:
Given the diameter of the circle as 20 cm as we know the diameter of the circle is equal to 2 X radius, so as the given circle has d = 20 cm so radius = diameter/ 2 = 20 cm/2 = 10 cm. We know area of the circle as A = π r², A = 3.14285 X 10 cm X 10 cm = 314.285 cm² and circumference of the circle is C = 2πr = 2 X 3.14285 X 10 cm = 62.857 cm.

Question 2.

What is the circumference of the circle? ___________
What is the radius? ___________
Answer:
The circumference of the circle is 14 π,
The radius is 7,

Explanation:
Given the area of the circle as 49π as we know area of the circle as A = π r²,
49 π = π X r X r, So r= square root of 49 so r = 7 and as circumference of the circle is C = 2πr = 2 X π X 7 = 14π.

Question 3.

What is the diameter of the circle? ___________
What is the area? ___________
What is the circumference? ___________
Answer:
The diameter of the circle is equal to 10 cm,
The area of the circle is 78.57125 cm²,
The circumference of the circle is 31.4285 cm,

Explanation:
As we know the diameter of the circle is equal to 2 X radius given r = 5 cm so as the given circle has d = 2 X 5 cm = 10 cm. As we know area of the circle as A = π r², A = 3.14285 X 5 cm X 5 cm = 78.57125 cm² and circumference of the circle is C = 2πr = 2 X 3.14285 X 5 cm = 31.4285 cm.

Question 4.

What is the circumference of the circle? ___________
What is the area of the circle? ___________
Answer:
The circumference of the circle is 37.7142 cm,
The area of the circle is  113.1426 cm²,

Explanation:
As we know the diameter of the circle is equal to 2 X radius given d = 12 cm so as the given circle has r = 12/2 cm = 6 cm. As we know circumference of the circle is C = 2πr = 2 X 3.14285 X 6 cm = 37.7142 cm and area of the circle as A = π r², A = 3.14285 X 6 cm X 6 cm = 113.1426 cm².

Question 5.
A bicycle wheel measures 25 inches in diameter. How far would a bike travel if the wheel went completely around 4 times?
____________ inches
Answer:
Far would a bike travel if the wheel went completely around 4 times is 78.57125 inches,

Explanation:
Given A bicycle wheel measures 25 inches in diameter. Far would a bike travel if the wheel went completely around 4 times as we know the diameter of the circle is equal to 2 X radius given d = 25 inches so as the given circle has r = 25/2 inches = 12.5 inches. As we know circumference of the circle is C = 2πr = 2 X 3.14285 X 12.5 inches = 78.57125 inches.

Question 6.
Brenda walks her dog along two circular routes laid out in the park. How much longer is Route B then route A? (Leave the answer in terms of pi.)

_____________ π km
Answer:
Longer is route B than route A is 10π km,

Explanation:
Given Brenda walks her dog along two circular routes laid out in the park. Much longer is Route B then route A is as circumference of route A is 2πr as r is 5 km so it is 2 X π X 5 km = 10π km and circumference of route B is 2πr as r is 10 km so it is 2 X π X 10 km = 20π km therefore longer is route B than route A is 20π km – 10π km = 10π km.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 21.2 Polygons to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 21.2 Polygons

Exercises

IDENTIFY

Indicate whether the figure is a polygon. Write “Polygon” or “Not a Polygon”.

Question 1.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 2.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 3.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 4.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 5.

Answer:
Not a Polygon,

Explanation:
The given shape is a not polygon as it is a closed 2-dimensional figure made up of not only line segments but also with curves.

Question 6.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 7.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 8.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Question 9.

Answer:
Polygon,

Explanation:
The given shape is a polygon as it is a closed 2-dimensional figure made up of line segments and it has more than 3 sides.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 21.1 Quadrilaterals to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 21.1 Quadrilaterals

Exercises

IDENTIFY

For each figure below, label as a square, rectangle, rhombus, trapezoid, or kite.

Question 1.

Answer:
Square,

Explanation:
We know if all the four sides are of the same length it is a square, as we have given shape have with all the four sides 4 in. so it is a square.

Question 2.

Answer:
Trapezoid,

Explanation:
As a trapezoid has two opposite sides that are parallel, but the sides are not the same
length and the other two sides of a trapezoid are not parallel, So given shape is trapezoid.

Question 3.

Answer:
Rectangle,

Explanation:
As the given shape has 2-dimensional figure with four right angles and the opposite sides are parallel and the same length so it is rectangle.

Question 4.

Answer:
Rectangle,

Explanation:
As the given shape has 2-dimensional figure with four right angles and the opposite sides are parallel with the same length a,a and b,bso it is rectangle.

Question 5.

Answer:
Trapezoid,

Explanation:
As a trapezoid has two opposite sides that are parallel, but the sides are not the same
length and the other two sides of a trapezoid are not parallel, So given shape is trapezoid.

Question 6.

Answer:
Square,

Explanation:
We know if all the four sides are of the same length it is a square, as we have given shape have with all the four sides equal so it is a square.

Question 7.

Answer:
Rhombus,

Explanation:
Given shape is rhombus with opposite sides are parallel and all the sides are the same length. Unlike a rectangle, a rhombus does not have four right angles.

Question 8.

Answer:
Kite,

Explanation:
Given shape is a kite as it is a quadrilateral which looks like a typical toy kite, so it is a kite, two of its angles are equal, Its longer two touching sides are equal in length and so
are shorter two touching sides.

Question 9.

Answer:
Rectangle,

Explanation:
As the given shape has 2-dimensional figure with four right angles and the opposite sides are parallel and the same length so it is rectangle.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 20.3 Right Triangles and Pythagorean Theorem to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 20.3 Right Triangles and Pythagorean Theorem

Exercises

SOLVE

Use the Pythagorean Theorem to determine the length of the missing side.

Question 1.
If side A is 6 and side B is 8, then side C (the hypotenuse) is _____________
Answer:
C = 10,

Explanation:
As side A is 6 and side B is 8, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A² + B² ) = square root of (6² + 8²)= square root of (36 + 64) = square root of 100 = 10.

Question 2.
If side A is 9 and side B is 9, then side C (the hypotenuse) is _____________
Answer:
C = 12.72,

Explanation:
As side A is 9 and side B is 9, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A² + B² ) = square root of (9² + 9²)= square root of (81 + 81) = square root of 162 , approximately equal to 12.72.

Question 3.
If side A is 10 and side B is 24, then side C (the hypotenuse) is _____________
Answer:
C = 26,

Explanation:
As side A is 10 and side B is 24, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A² + B² ) = square root of (10² + 24²)= square root of (100 + 576) = square root of 676 = 26.

Question 4.
If side A is 9 and side C (the hypotenuse) is 15, then side B is _____________
Answer:
B = 12,

Explanation:
As side A is 9 and side C (the hypotenuse) is 15, then side B is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. C = square root of (A² + B² ) therefore B = square root of (C^{2 –} A²)= square root of (225 – 81) = square root of 144 = 12.

Question 5.
If side B is 6 and the hypotenuse is 10, then side A is _____________
Answer:
A = 8,

Explanation:
As side B is 6 and the hypotenuse is 10, then side A is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A² + B² ) therefore A = square root of (C² – B²)= square root of (100 – 36) = square root of 64 = 8.

Question 6.
If side A is 12 and side B is 5, then side C (the hypotenuse) is _____________
Answer:
C = 13,

Explanation:
As side A is 12 and side B is 5, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A² + B² ) = square root of (12² + 5²) = square root of (144 + 25) = square root of 169 = 13.

Find the missing sides of the following pairs of similar right triangles:

Question 1.
HI = ___________ ft
KL = ___________ ft
JL = ____________ ft

Answer:
HI = 50 ft.,
KL = 30 ft.,
JL =  78 ft.,

Explanation:
For finding HI we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So HI = square root of (GI² – GH² ) = square root of (130² – 120²) = square root of (16,900 – 14,400) = square root of 2,500 = 50 ft.
Now finding KL let it be x as GHI and JKL are similar triangles that means the ratios of the sides are equal  HI/GH = KL/JK = 50/120 = x/72 cross multiplying for the unknown we get
120x = 50 X 72, x = 5 X 72/12 = 5 X 6 = 30 ft.
Now JL we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So HI = square root of (JK² + KL² ) = square root of (72² + 30²) = square root of (5,184 + 900) = square root of 6,084 = 78 ft.

Question 2.
LN = ___________ m
PQ = ___________ m
OQ = ____________ m

Answer:
LN = 25 ft., = 7.62 m,
PQ = 21 ft., = 6.4008 m,
OQ = 35 ft., = 10.668 m,

Explanation:
For finding LN we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So LN = square root of (LM² + MN² ) = square root of (20² + 15²) = square root of (400 + 225) = square root of 625 = 25 ft. As 1 foot is equal to 0.3048 meter,
so 25 X 0.3048 m = 7.62 m.
Now finding PQ let it be x as LMN and OPQ are similar triangles that means the ratios of the sides are equal  MN/LM = PQ/OP = 15/20 = x/28 cross multiplying for the unknown we get
20x = 15 X 28, x = 3 X 28/4 = 3 X 7 = 21 ft., = 21 X 0.3048 m = 6.4008 m.
Now OQ we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So OQ = square root of (OP² + PQ² ) = square root of (28² + 21²) = square root of (784 + 441) = square root of 1,225 = 35 ft., = 35 X 0.3048 m =  10.668 m.

Question 3.
RT = ___________ in.
VW = ___________ in.
UW = ____________ in.

Answer:
RT = 7.071 ft., = 84.852 in.,
VW = 20 ft., = 240 in.,
UW = 28.284 ft., = 339.400 in.,

Explanation:
For finding RT we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So HI = square root of (RS² + ST² ) = square root of (5² + 5²) = square root of (25 + 25) = square root of 50 = 7.071 ft. As 1 foot is equal to 12 inch we get 7.071 X 12 inch = 84.852 in, Now finding VW let it be x as RST and UVW are similar triangles that means the ratios of the sides are equal  ST/RS = VW/UV = 5/5 = x/20 cross multiplying for the unknown we get 5x = 5 X 20, x = 20 ft., = 20 X 12 in = 240 in.,
Now UW we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So UW = square root of (UV² + VW² ) = square root of (20² + 20²) = square root of (400 + 400) = square root of 800 = 28.284 ft., = 339.408 in.

Question 4.
If the two triangles are similar, then what is the length of the missing side?

Answer:
The length of the missing side is 16 ft.,

Explanation:
As given two triangles are similar triangles that means the ratios of the sides are equal, Let the missing side be x so 9/12 = 12/x cross multiplying for the unknown we get 9x = 12 X 12, x = 12 X 12/9 = 12 X 4 /3 = 4 X 4 = 16 ft.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 18.2 Line Segments and Rays to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 18.2 Line Segments and Rays

Exercises

SOLVE

Question 1.
Draw a diagram that illustrates a ray \(\overrightarrow{Y X}\).
Answer:

Explanation:
Drawn a diagram that illustrated a ray \(\overrightarrow{Y X}\) as shown above.

Question 2.
Identify two rays and two line segments in the figure.

Answer:
Two rays: \(\overrightarrow{XL}\), \(\overrightarrow{YM}[/latex,Two line segments: LX,YM

Explanation:
Identified two rays [latex]\overrightarrow{XL}\), \(\overrightarrow{YM}\) as rays means a part of a line that has one endpoint and goes on infinitely in only one direction. Two line segments LX, YM as in geometry, a line segment is bounded by two distinct points on a line also a line segment is part of the line that connects two points.

Question 3.
Identify three line segments in the diagram.

Answer:
Three line segments: AB, BC, CD,

Explanation:
The three line segments in the given diagram above are AB, BC, CD.

Question 4.
Draw two rays \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\).
Answer:

Explanation:
Drawn two rays \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) as shown above.

Question 5.
Identify the two line segments.

Answer:
CD, PN,

Explanation:
Identified the given line segments CD, PN as a line segment is bounded by two distinct points on a line. Or we can say a line segment is part of the line that connects two points.