We included **HMH Into Math Grade 8 Answer Key PDF** **Module 11 Review **to make students experts in learning maths.

## HMH Into Math Grade 8 Module 11 Review Answer Key

**Vocabulary**

**For Problems 1-3, choose the correct term from the Vocabulary box to complete each sentence.**

Vocabulary

hypotenuse

leg

Pythagorean Theorem

Pythagorean triple

Question 1.

The ____________ describes the relationship leg among the lengths of the sides of any right triangle.

Answer:

Pythagorean Theorem,

Explanation:

The Pythagorean Theorem describes the relationship leg among the lengths of the sides of any right triangle.

Question 2.

A _____________ of a right triangle is one of the sides that forms the right angle.

Answer:

hypotenuse,

Explanation:

A hypotenuse of a right triangle is one of the sides that forms the right angle.

Question 3.

A _____________ is a set of three positive integers that could be the side lengths of a right triangle.

Answer:

Pythagorean triple,

Explanation:

A Pythagorean triple is a set of three positive integers that

could be the side lengths of a right triangle.

Question 4.

Write the converse of this statement: If a triangle has a right angle, then it is a right triangle.

Answer:

Converse: If a triangle is right triangle then it has a right angle,

Explanation:

Given to write the converse of this statement: If a triangle has a right angle, then it is a right triangle so it is Converse: If a triangle is right triangle then it has a right angle.

**Concepts and Skills**

Question 5.

Which set of side lengths could form a right triangle?

(A) 5 cm, 5 cm, and 10 cm

(B) 6 cm, 7 cm, and 8 cm

(C) 8 cm, 15 cm, and 17 cm

(D) 9 cm, 12 cm, and 16 cm

Answer:

(C) 8 cm, 15 cm, and 17 cm and

(D) 9 cm, 12 cm, and 16 cm,

Explanation:

To find which set of side lengths could form a right triangle is checking with bit(A) 5 cm, 5 cm, and 10 cm we have (5)^{2} + (5)^{2} = 10^{2}, 25 + 25 = 100, as 50 ≠ 100, checking with bit(B) 6 cm, 7 cm, and 8 cm we have (6)^{2} + (7)^{2} = 8^{2}, 36 + 49 = 64, as 85 ≠ 64,

checking with bit(C) 8 cm, 15 cm, and 17 cm we have

(8)^{2} + (15)^{2} = 17^{2}, 64 + 225 = 289, as 289 = 289,

now checking with bit(D) 9 cm, 12 cm, and 16 cm we have (9)^{2} + (12)^{2} = 16^{2},

81 + 144 = 225 as 225 =225, therefore bits (C) and (D) set of side lengths could form a right triangle.

Question 6.

Use Tools The diagram represents a set of beams that form part of a bridge support. Label \(\overline{A B}\) and \(\overline{B D}\) with their lengths, rounded to the nearest foot. State what strategy and tool you will use to answer the question, explain your choice, and then find the answer.

Answer:

Explanation:

To find the distance of BD we calculate as BD^{2} = (14)^{2} – 7^{2},

BD^{2} = 196 – 49 = 147, So BD is equal to square root of 147 = 12.12 nearly 12 ft, Now the distance of AB is AB^{2} = (12)^{2} + 9^{2}, AB^{2} = 144 + 81 = 225, So AB is equal to square root of

225 = 15 ft, Used Tools the diagram represented a set of beams that form part of a bridge support.

Labelled \(\overline{A B}\) as 15 ft and\(\overline{B D}\) as 12 ft with their lengths and rounded to the nearest foot. Strategy used as both are right triangles is Pythagorean theorem.

**For Problems 7 and 8, determine the unknown side length of each right triangle to the nearest hundredth.**

Question 7.

What is the length of \(\overline{J L}\)?

__________ meters

Answer:

The length of \(\overline{J L}\) is 11 meters,

Explanation:

Given side length of triangle KJL as KL 14 m and KJ as 18 m as triangle is right triangle applying

Pythagorean theorem to find the length of JL as JL^{2} = (18)^{2} – 14^{2},

JL^{2} = 324 -196 = 128, So JL is equal to square root of 128 = 11.31 nearly 11 meters.

Question 8.

What is the length of \(\overline{S T}\)?

___________ inches

Answer:

The length of \(\overline{S T}\) is 22 in,

Explanation:

Given side length of triangle SRT as SR 8 in. and RT as 20 in. as triangle is right triangle applying

Pythagorean theorem to find the length of ST as ST^{2} = 8^{2} + 20^{2},

ST^{2} = 64 + 400 = 464, So ST is equal to square root of 464 = 21.54 nearly 22 in.

Question 9.

The steps shown can be used to prove the Pythagorean Theorem.

Explain how the steps prove the Pythagorean Theorem. Hint: Write expressions for the total area of the figures in Step 1 and in Step 4.

Answer:

c^{2} = a^{2} + b^{2} ,

Explanation:

Asking to prove the Pythagorean Theorem as a^{2} + b^{2} = c^{2}, So

Step 1: Drawn a figure using two squares one big square with side a we have area of

square as a^{2} and small square with side b and area as b^{2}, total area is a^{2} + b^{2}.

Step 2: Drawn two congruent right triangles inside the figure as shown above with hypotenuse as c,

Step 3: Rotated the two triangles as shown above,

Step 4: The resulting figure is a square of side length c.

Therefore the area of resulting sqaure is a^{2} + b^{2} = c^{2} which is equal to a^{2} + b^{2}. hence proved.

Question 10.

Which set of side lengths could form a right triangle?

(A) 3ft, 5ft, 12ft

(B) 5ft, 12ft, 13ft

(C) 12ft, 13ft, 16ft

(D) 13ft, 16ft, 24ft

Answer:

(B) 5ft, 12ft, 13ft,

Explanation:

To find which set of side lengths could form a right triangle is checking with bit(A) 3ft, 5ft, 12ft we have (3)^{2} + (5)^{2} = 12^{2}, 9 + 25 = 144, as 34 ≠ 144, checking with bit (B) 5ft, 12ft, 13ft we have

(5)^{2} + (12)^{2} = 13^{2}, 25 + 144 = 169, as 169 =169, checking with bit(C) 12ft, 13ft, 16ft we have

(12)^{2} + (13)^{2} = 16^{2}, 144 + 169 = 256, as 313 ≠ 256, now checking with bit(D) 13ft, 16ft, 24ft we have

(13)^{2} + (16)^{2} = 24^{2}, 169 + 256 = 576, as 425 ≠ 576, therefore bit (B) lengths could form a right triangle.

Question 11.

Meg is making a scale model of an Egyptian pyramid. The model is a right square pyramid as shown. To the nearest centimeter, what is the base length b of the model?

(A) 12 cm

(B) 16 cm

(C) 23 cm

(D) 48 cm

Answer:

(A) 12 cm,

Explanation:

Given Meg is making a scale model of an Egyptian pyramid. The model is a right square pyramid as shown above. The nearest centimeter is the base length b of the model is as it is right triangle applying Pythagorean theorem as19^{2} = b^{2} + 15^{2}, b^{2} = 361 – 225 = 136, So b is equal to square root of

11.66 nearly 12 cm which matches with bit (A).

**For Problems 12-14, determine the distance between the pair of points, to the nearest hundredth of a unit.**

Question 12.

A(-2, 1) and B(3, 5) ________________ units

Answer:

Distance between A and B is approximately 6 units,

Explanation:

Given A(-2, 1) and B(3, 5) the distance between the pair of points be x so applying Pythagorean theorem as x^{2} = (3 – (-2))^{2} + (5 – 1)^{2}, x^{2} = 25 + 16 = 41, So x is equal to square root of

41 = 6.40 nearly 6 units.

Question 13.

C(-5, -2) and D(0, -4) _______________ units

Answer:

Distance between C and D is approximately 5 units,

Explanation:

Given C(-5, -2) and D(0, -4) the distance between the pair of points be x so applying Pythagorean theorem as x^{2} = (0 – (-5))^{2} + (-4 – (-2))^{2}, x^{2} = 25 + 4 = 29, So x is equal to square root of

29 = 5.38 nearly 5 units.

Question 14.

E(7, 3) and F(-2, 6) _______________ units

Answer:

Distance between E and F is approximately 9 units,

Explanation:

Given E(7, 3) and F(-2, 6) the distance between the pair of points be x so applying Pythagorean theorem as x^{2} = (-2 – 7)^{2} + (6 – 3)^{2}, x^{2} = 81 + 9 = 90, So x is equal to square root of

90 = 9.48 nearly 9 units.