﻿ Converse of the Pythagorean Theorem Worksheet - Page 3 | Problems & Solutions # Converse of the Pythagorean Theorem Worksheet - Page 3

Converse of the Pythagorean Theorem Worksheet
• Page 3
21.
A Pythagorean triplet when multiplied by a positive integer $n$ is also a Pythagorean triplet. State whether the statement is true or false. a. True b. False

#### Solution:

Let a Pythagorean triplet be 9, 12 and 15.

Multiply each number by n.

Then the new set is 9n, 12n and 15n.

(15n)2 = (9n)2 + (12n)2
[Apply Pythagorean theorem.]

225n2 = 81n2 + 144n2
[Simplify.]

225n2 = (81 + 144)n2 = (225)n2

A Pythagorean triplet when multiplied by a positive integer n, the new set formed is also a Pythagorean triplet.

22.
The radius of the circle is 3 inches. Find the length of the sides of the square using the Pythagorean theorem.  a. 3$\sqrt{2}$inches b. 3 inches c. 6$\sqrt{2}$ inches d. 6 inches

#### Solution:

Given, the radius of the circle OB = 3 inches.

BD is the diameter, BD = 6 inches.

From right triangle BDC, BD2 = BC2 + CD2
[Apply Pythagorean theorem.]

BD2 = BC2 + BC2
[Since all sides of a square are equal,replace CD with BC.]

62 = 2BC2
[Substitute BD.]

36 = 2BC2
[Simplify.]

BC2 = 362 = 18
[Divide by 2 on both sides]

BC = 18 = 32
[Take square root of both sides.]

The length of side of square = 32 inches.

23.
The perimeter of a right triangle is 150 inches and the ratio of the lengths of the 2 legs is 5 : 12. Find the length of the hypotenuse. a. 13 inches b. 25 inches c. 60 inches d. 65 inches

#### Solution:

The two legs of a right triangle are in the ratio 5 : 12.

Let the length of two legs be 5k and 12k respectively.

Hypotenuse2 = (5k)2 + (12k)2
[Apply Pythagorean theorem.]

Hypotenuse2 = 25k2 + 144k2
[Apply exponents .]

= 169k2
[Simplify.]

Hypotenuse = 169k² = 13k
[Take square root of both sides.]

Perimeter of right triangle = 150 inches.

Perimeter of the right triangle = 5k + 12k + 13k = 30k.

30k = 150
[Equate perimeters]

k = 15030 = 5
[Divide by 30 on both sides.]

Hypotenuse = 13k = 13 × 5 = 65
[Substitute k.]

The length of hypotenuse = 65 inches.

24.
The length of the chord AB = 10 in. What is the radius of the circle?  a. 5 in b. 5$\sqrt{2}$ in c. 6$\sqrt{2}$ in d. 6 in

#### Solution:

Length of chord AB = 10 in.

AOB is a right triangle.

AB2 = OA2 + OB2
[Apply Pythagorean theorem.]

AB2 = 2OA2
[Since, the radius of the circle = OA = OB.]

OA2 = AB22
[Divide by 2 on both sides.]

OA2 = 1022
[Substitute AB.]

OA2 = 1002 = 50
[Apply exponents and simplify.]

OA = 50 = 52
[Find the positive square root.]

The radius of the circle is 52 in.

25.
State which of the following measures will form a right triangle?
(i) 9, 12, 15
(ii) 2, 4, $\sqrt{20}$
(iii) 3, 4, 6 a. (i) only b. (ii) only c. (iii) only d. both (i) and (ii)

#### Solution:

According to Pythagorean theorem, in a right triangle, square of the longest side = Sum of the squares of the other two sides.

In (i), 152 = 92 + 122

225 = 81 + 144 = 225

The measures 9, 12 and 15 will form a right triangle.

= 4 + 16 = 20
In (ii), (20)2 = 22 + 42

So, the measures 2, 4 and 20 will form a right triangle.

Applying Pythagorean theorem for (iii), 62 = 32 + 42

36 = 9 + 16 = 25, 36 is not equal to 25.

The measures 3, 4 and 6 cannot form a right triangle.

So, (i) and (ii) only will form right triangles.

26.
Which of the figures is a right triangle?  a. Only Figure 1 b. Only Figure 2 c. Only Figure 3 d. Both Figures 2 & 3

#### Solution:

According to Pythagorean theorem, in a right triangle, square of the longest side = Sum of the squares of other two sides.

Applying Pythagorean theorem for figure (1), 62 = 42 + 32 = 16 + 9

Since 36 ≠ 25, figure (1) is not a right triangle.

Applying Pythagorean theorem for figure (2), 102 = 62 + 82 = 36 + 64

100 = 100

So, figure (2) is a right triangle.

Applying Pythagorean theorem for figure (3), 42 = 22 + 22 = 4 + 4

Since 16 ≠ 8, figure (3) is not a right triangle.

Only figure (2) is the right triangle

27.
Jeff walked diagonally across a square garden with each side measuring 25 ft long (from one corner to the opposite corner). How far did he walk? a. 37.47 ft b. 33.23 ft c. 35.35 ft d. 35.47 ft

#### Solution:

Let s be the side of the square garden and d be the distance Jeff walked.

All the angles of a square are right angles. So, the diagonal of the square will be the hypotenuse of the right triangle formed by two adjacent sides and the diagonal.

d2 = s2 + s2
[Apply Pythagorean theorem.]

d2 = 252 + 252
[Substitute s = 25.]

d2 = 625 + 625 = 1250
[Apply exponents and simplify.]

d = 1250 = 35.35
[Take square root on both sides.]

The total distance Jeff walked is 35.35 ft.

28.
If one side of a right triangle is two times the other and the length of hypotenuse is 10 ft, then what are the measures of the two sides? a. 2 ft and 4 ft b. 3 ft and 5 ft c. 2$\sqrt{5}$ ft and 4$\sqrt{5}$ ft d. 3$\sqrt{5}$ ft and 4$\sqrt{5}$ ft

#### Solution:

Let one side of triangle be p. So, the other side of the right triangle is 2p.
[One side is two times the other.]

p2 + 4p2 = 102
[Apply Pythagorean theorem.]

5p2 = 100

p2 = 100 / 5 = 20
[Divide by 5 on both sides.]

p = 20 = 5 × 4 = 25
[Take square root on both sides.]

The other side = 2p = 2 × 25 = 45
[Substitute p.]

The two sides are 25 ft and 45 ft.