# Word Problems on Pythagorean Theorem | Pythagorean Theorem Problems with Answers

Check Word Problems On Pythagorean Theorem in the below sections. Refer Pythagorean Theorem Study Material along with the solutions and steps to solve them. Get various model questions on Pythagoreans theorem and improve your mathematical knowledge along with other math skills. Know the shortcuts, tips, and tricks to solve Pythagorean Theorem Problems. Assess your preparation standard taking the help of the Pythagorean Theorem Questions available and cross-check your solutions here.

## How to Solve Problems on Pythagorean Theorem?

Follow the simple steps listed here to solve problems related to the Pythagorean Theorem. They are along the lines

Step 1: Look at all the terms in the final equation

Step 2: Find out which right triangles contain those terms

Step 3: Start with those right triangles and apply the Pythagorean Theorem

## Pythagorean Theorem Word Problems

Problem 1:

A 35-foot ladder is leaning against the side of a building and its positioned such that the base of the ladder is 21 feet from the base of the building. Find the distance above the ground where the point of the ladder touches the building?

Solution:

Let the point above the ground = x

As given in the question,

Length of the foot ladder = 35

Base of the ladder = 21 feet

Let the equation be a² + b² = c²

(21)² + (x)² = (35)²

441 + x = 1225

Subtracting by 441 on both sides

x = 784

Squaring and both sides

x = √784

x = 28

28 feet far above the ground is the point where the ladder touches the building.

Problem 2:

Lainey runs a string of lights from the ground straight up a door frame that is 2.5 meters tall. Then they run the rest of the string in a straight line to a point on the ground that is 6 meters from the base of the door frame. There are 10 lights per meter of a string. How many total lights are on the string?

Solution:

As given in the question,

The length of the lights = 2.5 meters

The base of the door frame = 6 meters

No of lights per meter = 10

To find the total number of lights on a string, we write the equation

h² = (2.5)² + (6)²

h²= 6.25 + 36

h²= 42.25

h = √42.25

String Length = 2.5 + √42.25

Therefore, no of lights = 9m * 10 lights = 90 lights

Thus, the total lights on the string = 90 lights

Problem 3:

Fencing at a hardware store costs $16.97 per yard. How much would it cost to fence in a triangular-shaped yard that has a leg of 11 feet and another leg of 60 feet? Note: The yard is in the shape of a right triangle. Solution: Amount of fencing at hardware store =$16.97 per yard

Length of one leg = 11 feet

Length of another leg = 60 feet

As given the yard is in the shape of right triangle

We know the equation,

a² + b² = c²

(11)² + (60)² = c²

121 + 3600 = c²

3721 = c²

Squaring on both sides

√3721 = √c²

61 = c

The amount for a fence in a triangular-shaped yard = $61 Thus, the final solution is$61

Problem 4:

If a 34-foot ladder is placed against the top of a 30-foot building. How many feet will be at the bottom of the ladder from the bottom of the building?

Solution:

As given in the question,

Length of the ladder = 34 foot

Length of the building = 30 foot

Let the length of bottom of the ladder = x

a² + b² = c²

(30)² + x² = (34)²

900 + x² = 1156

x² = 256

Add square root on both sides

√x² = √256

x = 16 feets

Therefore, the bottom of the ladder from the bottom of the building will be 16 feets

Thus, the final solution is 16 feet

Problem 5:

A certain television is advertised as a 34-inch TV. If the width of the TV is 30 inches, how many inches tall is the TV?

Solution:

As given in the question,

Size of the TV = 34 inch

Width of the TV = 30 inch

Length of TV = x

To know the length of TV, we have to find the equation

a² + b² = c²

(30)² + x² = (34)²

900 + x² = 1156

Add square root on both sides

√x² = √1156

x = 34

TV is 34 inches tall.

Thus, the final solution is 34 inches

Problem 6:

Ramu starts driving north for 9 miles, then takes a right turn, and then he drives east for another 40 miles. At the end of driving, what is the distance of a straight line from the starting point?

Solution:

As given in the question,

Driving for north = 9 miles

Driving for east =40 miles

Let the straight line distance = x

The equation will be

a² + b² = c²

(9)² + (40)² = c²

81 + 1600 = c²

1681 = c²

Add square roots on both sides

√1681 = √c²

41 = c

Therefore, the straight line distance from the starting point is 41 miles

Thus, the final solution is 41 miles

Problem 7:

In a right-angled triangle, the hypotenuse square is equal to the sum of the squares of other two sides?

Solution:

Given: A right-angle triangle in which one of the sides is 90°

To prove: AC² = AB² + BC²

Construction: Draw BD ⊥ AC

∠A = ∠A (Common)

ΔADB ∼ ΔABC (By A-A Criteria)

AB² = AD * AC is the first equation

In Δ’s BDC and ABC

∠CDB = ∠ABC (Each 90°)

∠C = ∠C (Common)

ΔBDC ∼ ΔABC (By A-A Criteria)

DC/BC = BC/AC

BC² = AC * DC is the (2) equation

AB² = AD * AC is the (3) equation

Adding the equations (1) and (2)

AB² + BC² = AC (AD + DC)

AB² + BC² = AC (AC)

AB² + BC² = AC²

∴ Hence proved

Problem 8:

In the triangle, if the square of one side of the triangle is equal to the sum of the squares of another two sides of the triangle, then the angle that is opposite to the first side is a right angle triangle?

Solution:

Given: ΔABC such that AB² + BC² = AC²

To prove: ∠B = 90°

Construction: Draw ΔDEF such that DE = AB, EF = BC and ∠E = 90°

Since ΔDEF is right-angled,

By Pythagoras theorem

DE² + EF² = DF²

AB² + BC² = DF² (By const DE = AB and EF = BC)

AC² = DF² (Given AB² + BC² = AC²)

AC = DF

To prove: ∠B = 90°

In Δs ABC and DEF

AB = DE (By const)

BC = EF (By const)

AC = DF (Proved)

ΔABC ≅ ΔDEF (SSS)

∠B = ∠E = 90°

ΔABC is a right triangle

Problem 9:

Mary wants to cut across a rectangular lot rather than walk around it. Of the lot is 120 feet long and 50 feet wide. Mary walks diagonally across the lot, how many feet is the short cut?

Solution:

As given in the question,

Length of the lot = 120 feet

Width of the lot = 50 feet

Let the shortcut distance be x

As per the Pythagorean theorem,

a² + b² = c²

(50)² + (120)² = x²

2500 + 14,400 = x²

16,900 = x²

130 = x

Therefore, Mary walks 130 feet shortcut

Thus, the final solution is 130 feet

Problem 10:

The length of a living room is 2 feet less than twice its width. If the diagonal is 2 feet more than twice the width, find the dimensions of the room?

Solution:

Let the width of the living room = x

Given that

Length of the living room = 2 feet less than twice its width

Diagonal = 2 feet more than twice the width

From the given equations,

Width represents x² + (2x – 2)² = (2x + 2)²

Length represents x² = (2x + 2)² – (2x-2)²

From both the equations,

x² = (4x + 8x + 4) – (4x – 8x + 4)

x² = 16x

(x²-16x) = 0

x(x-16) = 0

x = 0, x=16

Therefore, the width = 16

Length = (2x – 2) = 2(16) – 2 = 30

Thus, the length of the living room = 30 feet

Width = 16 feet

Hence, the final solution is l = 30 feet and w = 16 feet

Problem 11:

A man goes 12m east and 9m north. Find the distance from the initial point?

Solution:

As given in the question,

The distance man goes to east = 12m

i.e., BC = 12m

The distance man goes to north = 9m

i.e., AB = 9m

To find the initial point (AC), apply the Pythagorean theorem

AC² = AB² + BC²

AC² = (9)² + (12)²

AC² = 81 + 144

AC² = 225

AC = √225

AC = 15

Therefore, he is 15m far from his initial point

Thus, the final solution is 15m

Problem 12:

Hari wants to hang a 7m long banner from the roof of her shop. The hooks for the strings are 10m apart. Hari wants the top of the banner to hang 1m below the roof. How long should each of the strings be?

Solution:

As per the question,

Length of the long banner = 7m

Hooks for strings = 10m apart

Length where the banner to hang below the roof = 1m

Let the length of the string = x

Hence, the equation will be

7 + 2x = 10

2x = 3

x = 1.5

Now, apply the Pythagorean theorem,

C² = (1)² + (1.5)²

C² = 1 + 2.25

C² = 3.25

C = √3.25

C = 1.80

Hence, each string should be 1.80m

Thus, the final solution is 1.80m

Problem 13:

Two kids are flying a kite with a string of 50 meters long. If the kids are 35 meters apart, how high is the kite off the ground?

Solution:

As given in the question,

Length of the string = 50 meters

The distance of kids apart = 35 meters

The height of the kite off the ground = b

Applying the Pythagorean theorem, we get

a² + b² = c²

(35)² + b² = (50)²

1225 + b² = 2500

b² = 1275

b = √1275

b = 35.70

Therefore, 35.70m is the kite off the ground

Thus, the final solution is 35.70 m

Problem 14:

A carpenter needs to add 2 braces to a barn door. If the door measures 12 * 16 feet, how much wood will he need for both braces?

Solution:

As given in the question,

No of braces = 2

Length of the door = 12 feet

The breadth of the door = 16 feet

To find the amount of wood for braces, we apply Pythagorean theorem

a² + b² = c²

(12)² + (16)² = c²

144 + 256 = c²

400 = c²

c = √400

c = 20

Amount of wood for both the braces = 2 * 20 = 40ft

Therefore, the carpenter requires 40 ft of wood for both the braces.

Thus, the final solution is 40 ft

Problem 15:

M and N are points on the sides RP and RQ respectively of triangle PQR right-angled at R. Prove that PN² + QM² = PQ² + MN²?

Solution:

PN² = PR² + RN² is the (1) equation

QM² = MR² + RQ² is the (2) equation

Adding the equation (1) and (2)

PN² + QM² = MR² + RQ² + PR² + RN²

PN² + QM² = (MR² + RN²) + (RQ² + PR²)

PN² + QM² = MN² + PQ²

∴ Hence, it is proved