Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.4 Interior and Exterior Angles to score better marks in the exam.

## Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.4 Answer Key Interior and Exterior Angles

### Math in Focus Grade 7 Chapter 6 Lesson 6.4 Guided Practice Answer Key

**Hands-On Activity**

Materials:

- scissors
- ruler

Explore The Sum Of The Angles In A Triangle

Work in pairs.

Step 1.

Draw and cut out a triangle. Label the three angles of the triangle as 1, 2, and 3.

Step 2.

Cut the triangle into three pieces, so that each piece contains a vertex.

Step 3.

Rearrange the cut pieces on a straight line. What do you notice about the sum of the measures of the three interior angles?

**Complete.**

Question 1.

Find the value of p.

Answer:

p° = 88°

Explanation:

∠RPQ + ∠QRP + ∠PQR = 180°

35° + p° + 57° = 180°

p° + 92° = 180°

p° = 180° – 92°

p° = 88°

**Complete.**

Question 2.

Triangle XYZ is an isosceles triangle and m∠XYZ = 55°. Find the value of x.

Answer:

∠XYZ + ∠XZY + ∠YXZ = 180°

55° + 55° + x° = 180°

110° +x° = 180°

x° = 180° – 110°

x° = 70°

**Find the value of each variable.**

Question 3.

Answer:

x° = 70°

Explanation:

x° + 62° = 132°

x° = 132° – 62°

x° = 70°

Question 4.

Answer:

x° = 76°

Explanation:

x° + 35°= 70° + 41°

x° + 35° = 111°

x° = 111° – 35°

x° = 76°

**Complete.**

Question 5.

Triangle ABC is an isosceles triangle. Find the measures of ∠1 and ∠2.

AB = AC

m∠ACB = m∠ABC

Answer:

∠1 = 255°, ∠2 = 95°

Explanation:

m∠ACB = m∠ABC

∠1 = 75° + 180°

∠1 = 255°

∠2 = 180° – 75°

∠2 = 95°

Question 6.

Triangle EFG is an isosceles triangle and \(\overleftrightarrow{E F}\) is parallel to \(\overleftrightarrow{H I}\). Find the value of x.

Answer:

x° = 120°

Explanation:

∠FEG = **∠**IGJ

= 30°

∠FGE = ∠FEG

= 30°

∠FEG + ∠FGE + ∠EFG = 180°

30° + 30° + x° = 180°

x° + 60° = 180°

x° = 180° -60°

x° = 120°

### Math in Focus Course 2B Practice 6.4 Answer Key

**Find the value of y.**

Question 1.

Answer:

y° = 61°

Explanation:

y° + 52° + 67° = 180°

y° + 119° = 180°

y° = 180° -119°

y° = 61°

Question 2.

Answer:

y° + 26° + 18° = 180°

y° + 44° = 180°

y° = 180° – 44°

y° = 136°

Question 3.

Answer:

20° + y° = 90° + 42°

20° + y° = 132°

y° = 132° -20°

y° = 112°

Question 4.

Answer:

90° + 27° + x° = 180°

117° + x° = 180°

x° = 180° -117°

x° = 63°

63° + 51° + y° = 180°

114° + y° = 180°

y° = 180° – 114°

y° = 66°

Question 5.

Answer:

y° = 122°

Explanation:

y° + 28° + 30° = 180°

y° + 58° = 180°

y° = 180° – 58°

y° = 122°

Question 6.

Answer:

x° = 57° , y° = 67°

Explanation:

x° + 123° = 180°

x° = 180° -123°

x° = 57°

56° + 57° + y° = 180°

y° + 113° = 180°

y° = 180° – 113°

y° = 67°

**Find m∠1 and m∠2 in each diagram.**

Question 7.

Answer:

62° + 90° + m∠1 = 18o° ∠s sum in triangle:

152° + m∠1 = 180° Simplify:

152° + m∠1 — 152° = 180° — 152° Subtract 152° from both sides:

m∠1 = 28° Simplify:

m∠2 = 90° + m∠1 Exterior ∠ of triangle:

m∠2 = 90° + 28° substitute

m∠2 = 118° simplify

m∠1 = 28°

m∠2 = 118°

Question 8.

Answer:

∠1 = 76°, ∠2 = 114°

Explanation:

∠1 + 28° + 76° = 180°

∠1 + 104° = 180°

∠1 = 180° – 104°

∠1 = 76°

76° + 38° + x° = 180°

114° + x° = 180°

x° = 180° – 114°

x° = 66°

x° + ∠2 = 180°

66° + ∠2 = 180°

∠2 = 180° – 66°

∠2 = 114°

Question 9.

Answer:

50° + 70° + m∠1 = 180° Corresponding ∠s and ∠s sum in triangle:

120° + m∠1 = 180° Simplify:

120° + m∠1 – 120° = 180° – 120° Subtract 120° from both sides:

m∠1 = 60° Simplify:

m∠2 = 50° + 70° Exterior ∠ of triangle:

m∠2 = 120° simplify

m∠1 = 60°

m∠2 = 120°

Question 10.

Answer:

70° = 43° + m∠1 Alternate interior ∠s

70° – 43° = 43° + m∠1 – 43° Subtract 43° from both sides:

m∠1 = 27° Simplify:

m∠2 + m∠1 + 43° = 180° ∠s sum of triangle:

m∠2 + 27° + 43° = 180°

m∠2 + 70° = 180° Simplify:

m∠2 + 70° — 70° = 180° — 70° Subtract 70° from both sides:

m∠2 = 110° Simplify:

m∠1 = 27

m∠2 = 110°

**Use an equation to find the value of y.**

Question 11.

Answer::

3y° = y° + 80° Exterior ∠ of triangle:

3y = y + 80 The equation for y is:

3y – y = y + 80 – y Subtract y from both sides:

2y = 80 Simplify:

\(\frac{2 y}{2}\) = \(\frac{80}{2}\) Divide by 2

y = 40 simplify

3y = y + 80

y = 40

Question 12.

Answer:

Let’s note by x° the measure of the congruent angles of the isosceles triangle

x° = 2y° Alternate interior ∠s

x° + 3y° = 180° Supplementary ∠s:

2y° + 3y° = 180° Substitute:

5y° = 180° Simplify:

\(\frac{5 y^{\circ}}{5}\) = \(\frac{180^{\circ}}{5}\) Divide by 5

y = 36 simplify

2y + 3y = 180

y = 36

**Find the value of x and name the type of triangle that is shown.**

Question 13.

Answer:

x° = 60°, x° = 60°

Explanation:

x° + x° + 60° = 180°

2x° + 60° = 180°

2x° = 180° – 60°

2x° = 120°

x° = 120 ÷ 2

x = 60°

Question 14.

Answer:

x° = 36°, x° = 36°

Explanation:

x°+ x°+ 108° = 180°

2x° + 108° = 180°

2x° = 180° – 108°

2x° = 72°

x° = 72÷2

x° = 36°

Question 15.

Answer:

x° = 8°

Explanation:

We all know that all the angles in a triangle add up to 180 degrees.

In the given figure, each triangle has 3 angles. Thus, we have the sum of three angles as shown:

A° + B° +C° = 180°

62°+ 90°+ x° = 180°

x°+ 172° = 180°

x° = 180° – 172°

x° = 8°

**Find the measure of each numbered angle.**

Question 16.

Answer:

m∠2 = 148° Corresponding ∠s:

m∠2 = m∠3 + 63° Vertical ∠s and exterior ∠ of triangle:

148° = m∠3 + 63° Substitute:

148° – 63° = m∠3 + 63° – 63° Subtract 63° from both sides:

m∠3 = 85° Simplify:

m∠1 + 63° = 180° supplementary ∠s

m∠1 + 63° – 63° = 180° – 63° subtract 63° from both sides

m∠1 = 117° simplify

m∠1 = 117°

m∠2 = 148°

m∠3 = 85°

Question 17.

Answer:

m∠1 + 140° = 180° Supplementary ∠s

m∠1 + 140° — 140° = 180° – 140° Subtract 140° from both sides:

m∠1 = 40° Simplify:

m∠2 + 15° + 40° = 180° Alternate interior ∠s and ∠s sum of triangle:

m∠2 + 55° = 180° Simplify:

m∠2 + 55° – 55° = 180° – 55° subtract 55° from both sides

m∠2 = 125° simplify

m∠1 = 40°

m∠2 = 125°

**Find the measure of each numbered angle.**

Question 18.

Answer:

We are given

m∠5 = 50° Corresponding ∠s

m∠3 = m∠5 = 50° Corresponding ∠s

m∠4 = 50° Alternate Interior ∠s

m∠4 + m∠2 + 77° = 180° substitute

m∠2 + 127° = 180° simplify

m∠2 + 127° – 127° = 180° — 127° Subtract 127° from both sides:

m∠2 = 53° Simplify:

114° = m∠1 + m∠4 Exterior ∠ of triangle:

114° = m∠1 + 50° Substitute:

114° – 50° = m∠1 + 50° + 50° Substitute 50° from both sides

m∠1 = 64° simplify

m∠1 = 64°

m∠2 = 53°

m∠3 = 50°

Question 19.

Answer:

we are given

m∠1 + 67° + 53° = 180° Sum of ∠s in triangle:

m∠1 + 120° = 180° Simplify:

m∠1 + 120° – 120° = 180° — 120° Subtract 120° from both sides:

m∠1 = 60° Simplify:

m∠3 = 67° Alternate interior ∠s

m∠2 = m∠1 + m∠3 vertical ∠s

m∠2 = 60° + 67° substitute

m∠2 = 127° simplify

m∠1 = 60°

m∠2 = 127°

Solve.

Question 20.

\(\overleftrightarrow{\mathrm{BE}}\) is parallel to \(\overleftrightarrow{\mathrm{FH}}\). Find the measure of A

∠CAD in terms of ∠1 and ∠2.

Answer:

m∠ADE = m∠2 Corresponding ∠s:

m∠ADE = m∠CAD + m∠1 Exterior ∠ of triangle:

m∠2 = m∠CAD + m∠1 Substitute:

m∠2 — m∠1 = m∠CAD + m∠1 — m∠1 subtract m∠1 from both sides:

m∠CAD = m∠2 – m∠1 Simplify:

m∠CAD = m∠2 – m∠1

Question 21.

Math Journal

Explain why each of the following statement is true.

a) A triangle cannot have two right angles.

Answer:

m∠A = m∠B = 90° a) Let’s assume that a triangle ABC has two right angles:

m∠A + m∠B + m∠C = 180° Sum of ∠s in triangle:

90° + 90° + m∠C = 180° Substitute:

180° + m∠C = 180° Simplify:

180° + m∠C – 180° = 180° – 180° Subtract 180° from both sides:

m∠C = 0° Simplify:

We got a false statement as the triangle does not exist in this case. Therefore we assumed a wrong statement So a triangle cannot have two right angles.

b) The interior angle measures of an isosceles triangle cannot be 96°, 43°, and 43°.

Answer:

Lets assume that a triangle has the measures of its angles 96°, 43°, 43°

96° + 43° + 43° = 182° We compute the sum of the angles’ measures:

But the sum of a triangle’s angles measures is 180°. We got 182° ≠ 180°, therefore the statement that we assumed is false. So the interior measures of a triangle cannot be 96°, 43°, 43°.

See proofs (sum of the measures of the triangle’s angles)

Question 22.

m∠1 = 2x°, m∠2 = (x – 5)°, and m∠3 = 100°. Use an equation to find the value of x and then find the measures of ∠1 and ∠2.

Answer:

∠1 = 80°, ∠2 = 35°

Explanation:

∠1 + ∠3 = 180°

2x° + 100° = 180°

2x° = 180° – 100°

2x° = 80°

x° = 80°÷2

x° = 40°

∠1 = 2×40 = 80°

∠2 = 40 – 5 =35°

**Use an equation to find the value of x.**

Question 23.

Answer:

x° = 56°

Explanation:

x° + 94°+ 30° = 180°

x° + 124° = 180°

x° = 180° – 124°

x° = 56°

Question 24.

Answer:

x° = 59°, 2x° = 118°

Explanation:

2x° + 62° = 180°

2x° = 180° – 62°

2x° = 118°

x° = 118°÷2

x° = 59°

x + 13 = 59° + 13° = 72°

**Use an equation to find the value of x.**

Question 25.

Answer:

x° = 50°, y° = 65°

Explanation:

y° + 115° = 180°

y° = 180° -115°

y° = 65°

65° + 65° + x° = 180°

x° + 130° = 180°

x° = 180° – 130°

x° = 50°

Question 26.

Answer:

x° + x° + 12° = 180°

2x° + 12° = 180°

2x° = 180° – 12°

2x° = 68°

x° = 68 ÷ 2

x° = 34°

**Brain@Work**

ABCD is a rhombus, and the measure of ∠BCD is 68°. BDE is a triangle where the measure of ∠BED is 36° and the measure of ∠BDE is 73°. Find m∠EBC. Show how you obtain your answer.

Answer:

m∠EBD + 36° + 73° = 180° Sum of ∠s in triangle:

m∠EBD + 109° = 180° Simplify:

m∠EBD + 109° — 109° = 180° — 109° Subtract 109° from both sides:

m∠EBD = 71° Simplify:

\(\overline{B C}\) = \(\overline{D C}\) ABCD rhombus

m∠CDB = m∠CBD ∆BCD isosceles:

68° + m∠CDB + m∠CBD = 180° Sum of ∠s in triangle:

68° + 2m∠CBD = 180°

68° + 2m∠CBD — 68° = 180° — 68° Subtract 68° from both sides:

2m∠CBD = 112° Simplify:

Divide by 2

m∠CBD = 56° Simplify:

m∠EBC = m∠EBD — m∠CBD Substitute:

m∠EBC = 71° — 56°

= 15°