﻿ Pythagorean Theorem Worksheet | Problems & Solutions

# Pythagorean Theorem Worksheet

Pythagorean Theorem Worksheet
• Page 1
1.
Pythagorean theorem is applicable for ________.
 a. obtuse triangles only b. any triangle c. right triangles only d. None of the above

#### Solution:

Pythagorean theorem is applicable only for right triangles.

Correct answer : (3)
2.
What is the length of the side PR of ΔPQR in the figure?

 a. $\sqrt{20}$ in. b. $\sqrt{50}$ in. c. 25 in. d. 5 in.

#### Solution:

In the figure, ΔPQR is a right triangle.

From the figure, PQ = 5 in. and QR = 5 in.

According to Pythagorean theorem, in a right triangle, square of the hypotenuse = sum of the squares of the other two sides.

In ΔPQR, PR is hypotenuse.

From ΔPQR, PR2 = PQ2 + QR2

PR2 = 52 + 52 = 25 + 25 = 50

PR = 50

The length of the side PR = 50 in.

Correct answer : (2)
3.
What is the length of the side PR in the right triangle?
PR

 a. √41 units b. 10 units c. √21 units d. 20 units

#### Solution:

In the figure, PQ = 5 units and QR = 4 units.

PQR is a right triangle and PR is the hypotenuse.

Applying Pythagorean theorem for ΔPQR, PR2 = PQ2 + QR2.

PR2 = 52 + 42.
[Substitute PQ = 5 and QR = 4.]

PR2 = 25 + 16 = 41.

PR = √41
[Take square root on both the sides.]

So, the length of the side PR is √41 units.

Correct answer : (1)
4.
What is the length of the third side of the triangle in the figure?

 a. $\sqrt{8}$ units b. 4 units c. $\sqrt{15}$ units d. $\sqrt{34}$ units

#### Solution:

One angle of the triangle is 90o. So, the triangle is a right triangle.

The side opposite to the right angle is hypotenuse.

Let p be the hypotenuse.

Applying Pythagorean theorem, p2 = 32 + 52

p2 = 9 + 25 = 34
[Evaluate powers and add.]

p = √34
[Take square root on both the sides.]

The third side of the triangle is √34 units.

Correct answer : (4)
5.
What are the values of $x$ and $y$ in the triangle?

 a. 24 units and 45 units b. 4 units and 15 units c. 2√5 units and 45 units d. √160 units and √41 units

#### Solution:

ΔBDC is a right triangle.

BC2 = BD2 + CD2
[Apply Pythagorean theorem.]

x2 = 122 + 42
[Substitute BC, BD and CD.]

x2 = 144 + 16 = 160
[Evaluate powers and simplify.]

x = √160
[Take square root both sides.]

ΔADC is a right triangle.

AC2 = AD2 + CD2
[Apply Pythagorean theorem.]

y2 = 52 + 42
[Substitute AC, AD and CD.]

y2 = 25 + 16 = 41
[Evaluate powers and simplify.]

y = √41
[Take square root both sides.]

x = √160 units and y = √41 units.

Correct answer : (4)
6.
Which of the following measures make a right triangle?
 a. 7 cm, 24 cm, 25 cm b. 3 cm, 4 cm, 7 cm c. 8 cm, 24 cm, 25 cm d. 5 cm, 12 cm, 14 cm

Correct answer : (1)
7.
Find the length of the side PR in the ΔPQR shown.

 a. 25 in. b. $\sqrt{20}$ in. c. 5 in. d. $\sqrt{50}$ in.

Correct answer : (4)
8.
P is a point in the interior of the rectangle ABCD. AP = 15 units, CP = 5 units and DP = 8 units. What is the value of $\stackrel{‾}{\mathrm{BP}}$?

 a. 9.3 units b. 12 units c. $\sqrt{264}$ units d. $\sqrt{186}$ units

#### Solution:

MN is drawn || to AB through P.

BM = AN = a, CM = DN = b, PM = c, PN = d
[Figure.]

BPÃ‚Â² = MPÃ‚Â² + BMÃ‚Â² = a² + c²
[Pythagorean Theorem.]

CPÃ‚Â² = MPÃ‚Â² + CMÃ‚Â² ; 25 = c² + b²
[Pythagorean Theorem.]

DPÃ‚Â² = DNÃ‚Â² + PNÃ‚Â² ; 64 = b² + d²
[Pythagorean Theorem.]

APÃ‚Â² = ANÃ‚Â² + PNÃ‚Â² ; 225 = a² + d²
[Pythagorean Theorem.]

a² + c² = APÃ‚Â² + CPÃ‚Â² - DPÃ‚Â²
[Simplify using steps 5, 6 and 7.]

= 225 + 25 - 64 = 186.

So, BP = 186 units
[Steps 4 and 9.]

Correct answer : (4)
9.
The perimeter of a right triangle is 48 units. The sum of the squares of all the sides is 800 sq.units. What is the area of the triangle?
 a. 116 sq.units b. 96 sq.units c. 106 sq.units d. 111 sq.units

#### Solution:

Let the legs of the right triangles be a & b and the hypotenuse, c

Sum of squares = a2+b2+c2 = 800

2c2 = 800 Þ c2 = 400
[Pythagorean theorem.]

c = 20
[Simplify.]

a + b + c = 48
[Given.]

a + b =28

a2+b2 + 2 × (ab) = 784
[Squaring on both sides.]

2ab = 384
[Substitute the value of a2 + b2.]

Area of the triangle = ab2 = 96 sq.units

Correct answer : (2)
10.
The perimeter of a right triangle is 108 cm and the sum of the squares of its sides is 4050 sq.cm. Find the length of the smallest side.
 a. 36 cm b. 44.36 cm c. $\frac{75}{2}$cm d. 27 cm

#### Solution:

Let a, b are the lengths of legs and d be the hypotenuse.

a + b + d = 108 and a2 + b2 + d2 = 4050

d2 = a2 + b2
[Apply Pythagorean theorem.]

2d2 = a2 + b2 + d2 = 4050
[Add d2 on both sides.]

d2 = 2025
[Divide by 2 on both sides.]

d = 45
[Take square root of both sides.]

a + b + d = 108 a + b = 63
[Substitute d = 45.]

a2 + b2 + 2ab = 3969
[Square on both sides.]

2025 + 2ab = 3969 2ab = 1944
[Substitute a2 + b2 = 2025.]

ab = 972
[Divide by 2 on both sides.]

a + b = 63 and ab = 972
[From step 7 and step 10.]

So, a = 27, b = 36
[Solve for a and b.]

So, the length of the smallest side is 27 cm.

Correct answer : (4)

*AP and SAT are registered trademarks of the College Board.