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McGraw-Hill Math Grade 7 Answer Key Lesson 26.6 Volume of Solid Figures

Exercises

CALCULATE VOLUME

Question 1.

Answer:
Volume = 175π cu in.

Explanation:
Volume (V) = πr²h
V = π x 5 x 5 x 7
V = 175π cu in.

Question 2.

Answer:
Volume = 1125π cu in.

Explanation:
Volume (V) = πr²h
V = π x 15 x 15 x 5
V = 1125π cu in.

Question 3.

Answer:
Volume = 100 cu in.

Explanation:
Volume of a cuboid = (length × width × height) cubic units.
Volume (V) = (l × w × h) cubic units.
V = 10 x 5 x 2
V = 100 cu in.

Question 4.

Answer:
Volume = 154 cu in.

Explanation:
Volume of a cuboid = (length × width × height) cubic units.
Volume (V) = (l × w × h) cubic units.
V = 4 x 3.5 x 11
V = 154 cu in.

Question 5.

Answer:
Volume = 108 cu in.
Explanation:
Volume of a Triangular Prism =(1/2) (length × width × height) cubic units.
B = length x base
Volume of a Triangular Prism =(1/2) (length × width × height) cubic units.
Volume (V) =(1/2) (B × H) cubic units.
V = (1/2) x 3 x 6 x 12
V = 108 cu in.

Question 6.

Answer:
Volume = 17 cu in.
Explanation:
B = length x base
Volume of a Triangular Prism =(1/2) (length × width × height) cubic units.
Volume (V) =(1/2) (B × H) cubic units.
V = (1/2) x 1 x 2 x 17
V = 17 cu in.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.5 Surface Area of Solid Figures existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.5 Surface Area of Solid Figures

Exercises

CALCULATE SURFACE AREA

Question 1.

Answer:
SA = 54 sq in.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 3 x 3 + 2 x 3 x 3 + 2 x 3 x 3
SA = 18 + 18 + 18
SA = 54 sq in.

Question 2.

Answer:
SA = 104 sq m.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 6 x 5 + 2 x 5 x 2 + 2 x 2 x 6
SA = 60 + 20 + 24
SA = 104 sq in
Question 3.

Answer:
SA = 450.325 sq in.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 7.25 x 10 + 2 x 10 x 8.85 + 2 x 8.85 x 7.25
SA = 145 + 177 + 128.325
SA = 450.325 sq in

Question 4.

Answer:
SA = 125.46 sq ft.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 4.2 x 1.5 + 2 x 1.5 x 9.9 + 2 x 9.9 x 4.2
SA = 12.6 + 29.7 + 83.16
SA = 125.46 sq ft.

Question 5.

Answer:
SA = 856 sq cm.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 10 x 12 + 2 x 12 x 14 + 2 x 14 x 10
SA = 240 + 336 + 280
SA = 856 sq cm.
Question 6.

Answer:
SA = 862 sq m.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 11 x 12 + 2 x 12 x 13 + 2 x 13 x 11
SA = 264 + 312 + 286
SA = 862 sq m.

Question 7.

Answer:
SA = 478 sq m.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 9 x 7 + 2 x 7 x 11 + 2 x 11 x 9
SA = 126 + 154 + 198
SA = 478 sq m

Question 8.

Answer:
SA = 114 sq in.

Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 8 x 3 + 2 x 3 x 3 + 2 x 3 x 8
SA = 48 + 18 + 48
SA = 114 sq in

Question 9.

Answer:
SA = 370 sq ft.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 5 x 25 + 2 x 25 x 2 + 2 x 2 x 5
SA = 250 + 100 + 20
SA = 370 sq ft

Keep your answer in a form of pi.

Question 10.

Answer:
SA = 60π sq cm.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 3 (7 + 3)
SA = 60π sq. units

Question 11.

Answer:
SA = 170π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 5( 12 + 5)
SA = 170π sq in.

Question 12.

Answer:
SA = 252π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 9 ( 5 + 9)
SA = 252π sq in.

Question 13.

Answer:
SA = 275π sq m.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x11(1.5 + 11)
SA = 275π sq m.

Question 14.

Answer:
SA = 96.3π sq ft.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 4.5(6.2 + 4.5)
SA = 96.3π sq ft.

Question 15.

Answer:
SA = 352π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 8(14+ 8)
SA = 352π sq in.

Question 16.

Answer:
SA = 84π sq cm.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 6( 1 + 6)
SA = 84π sq cm.

Question 17.

Answer:
SA = 138.125π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 4.25 (12 + 4.25)
SA = 138.125π sq in.

Question 18.

Answer:
SA = 600π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x10 (20 + 10)
SA = 600π sq in.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.4 Circles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.4 Circles

Exercises

IDENTIFY

Question 1.
What is the radius of the circle?

Answer:
5,
Explanation:
A radius is a line segment that begins at a circle’s origin and extends to tis circumference.
in the circle , all radii are equal in the length.
So, radius r = 5 units.

Question 2.
What is the radius of the circle?

Answer:
5 cm,
Explanation:
A radius is a line segment that begins at a circle’s origin and extends to tis circumference.
In the circle , all radii are equal in the length.
diameter d = 10 cm
radius r = 5 cm.

Question 3.
Identify the 2 radii and the one chord below.

Answer:
\(\overline{AX}\) and \(\overline{XB}\) radii, \(\overline{AB}\) chord.
Explanation:
line \(\overline{AX}\) and \(\overline{XB}\) are radii lines which are passing though center point.
\(\overline{AB}\) is chord, which is not touching center of the circle.
A line segment that has both the end points on the circumference of the circle is called a chord.

Question 4.
Identify the radius and chord in the figure below.

Answer:
\(\overline{VY}\) radius,
\(\overline{WX}\) is chord.
Explanation:
A radius is a line segment that begins at a circle’s origin and extends to tis circumference.
In the circle , all radii are equal in the length.
\(\overline{VY}\) is a radius line which is passing though center,
and \(\overline{WX}\) is chord.
A line segment that has both the end points on the circumference of the circle is called a chord.

Question 5.
What are the 5 chords formed by inscribing the pentagon inside of the circle below?

Answer:
\(\overline{XY}\), \(\overline{YZ}\), \(\overline{ZV}\), \(\overline{VW}\) and \(\overline{WX}\).
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
pentagon has five chords.
\(\overline{XY}\), \(\overline{YZ}\), \(\overline{ZV}\), \(\overline{VW}\) and \(\overline{WX}\) are chords of the circle.

Question 6.
Using the letters provided, can the diameter in the figure below be named? Explain your answer.

Answer:
No, \(\overline{CD}\) does not go through center point O.
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
\(\overline{AB}\) and \(\overline{CD}\) are chords of the circle

Question 7.
How many ways can you describe the radius in this circle?

Answer:
\(\overline{EA}\), \(\overline{AE}\), \(\overline{EC}\), \(\overline{EB}\), \(\overline{CE}\), and \(\overline{BE}\)
Explanation:
\(\overline{EA}\), \(\overline{AE}\), \(\overline{EC}\), \(\overline{EB}\), \(\overline{CE}\), and \(\overline{BE}\) are radii of the circle,
which are originating at center of the circle and end at on the circumference of the circle.

Question 8.
Which two lines in the circle below are chords?

Answer:
\(\overline{AB}\) and \(\overline{CD}\) are chords of the circle.
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
\(\overline{AB}\) and \(\overline{CD}\) are chords of the circle.

Question 9.
Identify the chord and the diameter below.

Answer:
\(\overline{RA}\) is diameter of the circle,
\(\overline{QA}\) is the chord of the circle.
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
\(\overline{RA}\) is diameter of the circle,
\(\overline{QA}\) is the chord of the circle.

Question 10.
Calculate the area of the circle below. Leave your answer in the form of pi.

Answer:
100π in
Explanation:
Area of the circle = πr²
radius of the circle is 10 in
A = π x 10 x 10
A = 100π in

Calculate the circumference and the area.

Question 11.

Answer:
Circumference = __________________
Area = _____________
Answer:
Circumference = 10π  units.
Area = 25π sq units.
Explanation:
Circumference of the circle C = 2 π r
C = 2 π 5
C = 10π units
Area of the circle = πr²
radius of the circle is 10 in
A = π x 5 x 5
A = 25π sq units.

Question 12.

Circumference = ____________
Area = _____________
Answer:
Circumference = 14π units.
Area = 49π sq units.
Explanation:
Circumference of the circle C = 2 π r
C = 2 π 7
C = 14π units
Area of the circle = πr²
radius of the circle is 7 in
A = π x 7 x 7
A = 49π sq units.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.3 Polygons existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.3 Polygons

Exercises

IDENTIFY

Label the polygons.

Question 1.

Answer:
Heptagon.
Explanation:
Hepta means seven and gon means sides.
A Heptagon is a polygon with seven sides and seven angles.
It has seven straight sides and seven corners or vertices.

Question 2.

Answer:
Octagon.
Explanation:
Octa means eight and gon means sides.
So, octagon consists of 8 sides.
It has eight angles and eight corners.

Question 3.

Answer:
Pentagon.
Explanation:
Penta denotes five and gon denotes angle.
A pentagon is a simple polygon, which has five sides and five angles.

Question 4.

Answer:
Hexagon.
Explanation:
Hexa means six and gona means angles.
A hexagon is a closed two-dimensional polygon with six sides.
Hexagon has 6 vertices and 6 angles.

Question 5.

Answer:
Hexagon.
Explanation:
Hexa means six and gona means angles.
A hexagon is a closed two-dimensional polygon with six sides.
Hexagon has 6 vertices and 6 angles.

Question 6.

Answer:
Octagon.
Explanation:
Octa means eight and gon means sides.
So, octagon consists of 8 sides.
It has eight angles and eight corners.

Question 7.

Answer:
Heptagon.
Explanation:
Hepta means seven and gon means sides.
A Heptagon is a polygon with seven sides and seven angles.
It has seven straight sides and seven corners or vertices.

Question 8.

Answer:
Pentagon.
Explanation:
Penta denotes five and gon denotes angle.
A pentagon is a simple polygon, which has five sides and five angles.

Question 9.
Are these two figures congruent? Why or why not?

Answer:
Yes, sides and angle measures are same.
Explanation:
Two geometric figures are said to be congruent,
if they have same size and shape.
The mirror image of one shape is same as the other.

Question 10.
Are these two figures congruent? Why or why not?

Answer:
No, bases are of different lengths.
Explanation:
Two geometric figures are said to be congruent,
if they have same size and shape.
The above given images are not congruent,
As, their bases are different.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.2 Quadrilaterals existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.2 Quadrilaterals

Exercises

IDENTIFY

Label the shape as: square, rectangle, rhombus, trapezoid, or kite.

Question 1.

Answer:
Square,
Explanation:
A square is closed, two-dimensional shape with 4 equal sides.

Question 2.

Answer:
Kite,
Explanation:
A flat shape with 4 straight sides that has two pairs of sides,
which are two adjacent sides and angles are equal in length.

Question 3.

Answer:
Rectangle,
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees.
The two sides at each corner or vertex meet at the right angles.
The opposite sides of the rectangle are equal in length which makes it different from a square.

Question 4.

Answer:
Rhombus,
Explanation:
Rhombus is a quadrilateral with all equal sides.
Since opposite sides of a parallelogram are equal.
So, Rhombus is a special type of a parallelogram whose all sides are equal.

Question 5.

Answer:
Kite,
Explanation:
A flat shape with 4 straight sides that has two pairs of sides,
which are two adjacent sides that are equal in length.

Question 6.

Answer:
Rhombus,
Explanation:
Rhombus is a quadrilateral with all equal sides.
Since opposite sides of a parallelogram are equal.
So, Rhombus is a special type of a parallelogram whose all sides are equal.

Question 7.

Answer:
Trapezoid,
Explanation:
A trapezoid is a flat closed shape having 4 straight sides,
with one pair of parallel sides.

Question 8.

Answer:
Rectangle,
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees.
The two sides at each corner or vertex meet at the right angles.
The opposite sides of the rectangle are equal in length which makes it different from a square.

Question 9.

Answer:
Square,
Explanation:
A square is closed, two-dimensional shape with 4 equal sides and angles.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.1 Triangles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.1 Triangles

Exercises

IDENTIFY

Label the triangle as acute, right, or obtuse.

Question 1.

Answer:
Acute Angle,
Explanation:
An angle which is measuring less than 90 degrees is called an acute angle.
This angle is smaller than the right angle (which is equal to 90 degrees).
For example, ∠30^o, ∠45^o, ∠60^o, 75^o, etc. are all acute angles.

Question 2.

Answer:
Obtuse Angle,
Explanation:
The definition of an obtuse angle in geometry states that,
an angle whose measure is greater than 90° and less than 180° is called an obtuse angle.

Question 3.

Answer:
Right Angle,
Explanation:
x° = 60°
A right angle is an angle of 90°.
When two rays intersect and form a 90˚ angle or are perpendicular to each other,
at the intersection, they are said to form a right angle.

Question 4.

Answer:
Right Angle,
Explanation:
A right angle is an angle of 90°.
When two rays intersect and form a 90˚ angle or are perpendicular to each other,
at the intersection, they are said to form a right angle.

Question 5.

Answer:
Acute Angle,
Explanation:
An angle which is measuring less than 90 degrees is called an acute angle.
This angle is smaller than the right angle (which is equal to 90 degrees).

Question 6.

Answer:
Obtuse angle,
Explanation:
The definition of an obtuse angle in geometry states that,
an angle whose measure is greater than 90° and less than 180° is called an obtuse angle.

Label the triangle as equilateral, isosceles, or scalene.

Question 7.

Answer:
Isosceles,
Explanation:
An isosceles triangle is a triangle with two equal sides.
In the figure above, the two equal sides have same length.
An Isosceles triangle therefore has both two equal sides and two equal angles.

Question 8.

Answer:
Equilateral,
Explanation:
An equilateral triangle is a triangle with all three sides of equal length.
Total sum of triangle = 180°.
Given 2 angles = 60° + 60° = 120°.
Third angle = 180° – 120° = 60°.
Hence, the given angle is Equilateral triangle.

Question 9.

Answer:
Scalene Triangle,
Explanation:
All angles of a scalene triangle are unequal, all are of different size.
A scalene triangle has no line of symmetry.
The angle opposite to the longest side would be the greatest angle and vice versa.
All sides of the given scalene triangle are unequal.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 25.2 Types of Angles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 25.2 Types of Angles

Exercises

IDENTIFY

Question 1.
What is the measure of angle Q?

Answer:
50°

Explanation:
The sum of the two angles are 180 degrees.
So, ∠Q = 90° – 40° = 50°

Question 2.
What is the measure of angle Z?

Answer:
103°
Explanation:
The sum of two angles are 180°
∠A = 77°
So, ∠Z = 180° – 77° = 103°

Question 3.
Identify the supplementary angles in the figure below.

Answer:
∠T and ∠V; ∠T and ∠S; ∠S and ∠X; ∠X and ∠V
Explanation:
Two angles are called supplementary when their measures add up to 180 degrees.

Question 4.
The measure of ∠G is given. Determine the measure of the remaining angles.

Answer:
∠F = 120°; ∠E = 60°; ∠H = 120°
Explanation:
Two angles are called supplementary when their measures add up to 180 degrees.
∠G = 60°
So, ∠F = 180° – 60° = 120°
∠E = 180° – 120° = 60°
∠H is opposite of ∠F
So, ∠H = 120°

SOLVE

Question 1.
From this figure, give examples of two complementary, two supplementary, and two vertical angles.

Complementary _______________
Supplementary _____________
Vertical ______________
Answer:
Complementary angles are ∠BGC and ∠BGA; ∠FGE and ∠DGE.
Supplementary angles are ∠BCG and ∠BGF; ∠AGC and ∠AGF; ∠BGA and ∠AGE.
Vertical angles are ∠BGC and ∠FGE; ∠DGC and ∠AGF; ∠AGB and ∠DGE.
Explanation:
Two angles are called complementary when their measures add to 90 degrees.
Two angles are called supplementary when their measures add up to 180 degrees.
Vertical angles are angles that are opposite of each other when two lines cross.
Vertical angles are always congruent.

Question 2.
For the figure below, list all complementary and supplementary angles.

Complementary _______________
Supplementary _____________
Answer:
Complementary angles are ∠DFE and ∠CFD; ∠CFB and ∠AFB.
Supplementary angles are ∠AFB and ∠EFB; ∠AFC and ∠CFE; ∠DFE and ∠DFA.
Explanation:
Two angles are called complementary when their measures add to 90 degrees.
Two angles are called supplementary when their measures add up to 180 degrees.

Question 3.

Are angle DOB and angle DOC complementary? Explain.
Answer:
Yes, they form a right angle.
Explanation:
When we add two angles, there sum should be 90°,
∠DOB + ∠DOC = ∠COB
Given, ∠DOB = 40° and ∠COB = 90°
∠DOC = 90° – 40° = 50°
So, the given angles are complementary angles.

Question 4.
Identify the measure of the angles in this figure.

COD ______________
COF ______________
FOY ______________
YOX ______________
FOX ______________
Answer:
COD = 50°
COF = 90°
FOY = 40°
YOX = 50°
FOX = 90°

Explanation:
The measure of angles formed by the two rays at a common vertex.
Angles are measured in degrees.
When we add two angles, there sum should be 90°,
∠DOB + ∠DOC = ∠COB
Given, ∠DOB = 40° and ∠COB = 90°
∠DOC = 90° – 40° = 50°
∠COF = 90°
∠DOB = 40° and ∠DOC = 50°
∠FOY = 40° and ∠YOX = 90°
∠FOY + ∠YOX = ∠FOX
40° + 50° = 90°

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 25.1 Measuring Angles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 25.1 Measuring Angles

Exercises

IDENTIFY

Label the angles as acute, obtuse, or right.

Question 1.

Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 2.

Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 3.

Answer:
Right angle,
Explanation:
If the angle formed between two rays is exactly 90° then it is called a right angle.

Question 4.

Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 5.

Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 6.

Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 7.

∠DEF = _____________
Answer:
Right angle,
Explanation:
If the angle formed between two rays is exactly 90° then it is called a right angle.

Question 8.

∠RST = _____________
Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 9.

∠LMN = _____________
Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Question 10.

∠QRS = ______________
Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 11.

∠ABC = ______________
Answer:
Acute,
Explanation:
If two rays intersect at a vertex, forming an angle that is less than 90° is known as acute.

Question 12.

∠GHI = ______________
Answer:
Obtuse,
Explanation:
Any angle that is greater than 90° but less than 180° is known as obtuse angle.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 24.2 Line Segments and Rays existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 24.2 Line Segments and Rays

Exercises

IDENTIFY

Question 1.
How many line segments are there if you connect all the vertexes in this figure? Some, but not all, have been connected for you.

Answer:
30 lines segments,
Explanation:

there are 5 line segments from point A to B,C,D,E AND F.
6 points and 5 line segments,
6 x 5 = 30
Total 30 line segments can be constructed to connect all the vertexes in the above figure.

Question 2.
Let A stand for the town Augusta, and B stand for the town Bar Harbor. Using these two letters, what would be the ray if you were traveling from Bar Harbor through Augusta?
Answer:
\(\overline{BA}\),
Explanation:
Point A stand for the town Augusta, and B stand for the town Bar Harbor.
Using these two letters a ray from Bar Harbor through Augusta can be drawn.

Question 3.
Name all the possible line segments in this figure.

Answer:
\(\overline{BO}\), \(\overline{OD}\), \(\overline{BD}\), \(\overline{AB}\),
\(\overline{AO}\), \(\overline{OC}\), \(\overline{CB}\), \(\overline{BC}\),
\(\overline{OB}\), \(\overline{DO}\), \(\overline{DB}\), \(\overline{BA}\), \(\overline{OA}\), \(\overline{CO}\).
Explanation:
A line segment is part of a line that has two endpoints and is finite in length.
A ray is a line segment that extends indefinitely in one direction.
The following are all possible line segments, which can be drawn from O to B and O to D and all…
\(\overline{BO}\), \(\overline{OD}\), \(\overline{BD}\), \(\overline{AB}\),
\(\overline{AO}\), \(\overline{OC}\), \(\overline{CB}\), \(\overline{BC}\),
\(\overline{OB}\), \(\overline{DO}\), \(\overline{DB}\), \(\overline{BA}\), \(\overline{OA}\), \(\overline{CO}\).

Question 4.
At the right is a drawing of an airport runway. Name all the possible rays that could be used to describe the different ways a plane could take off. For points, use the numbers at the end of the runways: 15, 24, 8 and 33. An example of one is ray 15 – 33. Name the other three.

Answer:
Ray 33-15, Ray 24-8, Ray 8-24, Ray 15-33
Explanation:

A Ray from DC and CD Ray 33-15, Ray 15-33
A Ray from AB and BA Ray 24-8, Ray 8-24

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 24.1 Points and Lines existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 24.1 Points and Lines

Exercises

IDENTIFY

Question 1.
List the points located in the figure below.

Answer:
A, B, C, D
Explanation:
The below figure shown the points located A, B, C and D are vertex.
A point where two or more line segments join together are known as vertex.

Question 2.
How many different points are located in the figure below?

Answer:
8 points
Explanation:
A point where two or more line segments join together are known as vertex.

8 points are located in the above figure.
Each point is one vertex of the above figure.

Question 3.
Do the figures for exercises 1 and 2 contain any lines? Why or why not?
Answer:
No, line must stretch infinitely in both directions.
Explanation:
The above solid figures given in 1 and 2 dose not contain any lines.

Question 4.
List all possible names for the line pictured below.

Answer:
\(\overleftrightarrow{XY}\), \(\overleftrightarrow{YX}\), \(\overleftrightarrow{ZX}\), \(\overleftrightarrow{XZ}\), \(\overleftrightarrow{ZY}\) and \(\overleftrightarrow{YZ}\),
Explanation:

The above figure is a line with two points on the line X and Y with a center point Z and pointed.
the possible names for the line pictured are written below,
\(\overleftrightarrow{XY}\), \(\overleftrightarrow{YX}\), \(\overleftrightarrow{ZX}\), \(\overleftrightarrow{XZ}\), \(\overleftrightarrow{ZY}\), \(\overleftrightarrow{YZ}\).