Pythagorean Theorem (Word Problems)

Check out how to solve the different types of word problems on the Pythagorean Theorem. But first, Pythagoras Theorem can be useful for solving the word problems step-by-step when we know the length of both sides of the right angle and we need to check out the length of the remaining side. There are three possibilities of word problems on Pythagoras Theorem

1. For finding the hypotenuse, where the perpendicular side and base are known.
2. For finding the base length, where the perpendicular and hypotenuse are known.
3. For finding the perpendicular side, where the base and hypotenuse are known.

Example 1 – If the square of the hypotenuse of an isosceles right-angled triangle is 128 square cm, find out the length of each side.

Note: for an isosceles right triangle, the legs are the same length.

pythagorean 1

We get C2 = A2 + A2
128 = 2A2
128 / 2 = A2
64 = A2

Thus, A = 8

Thus, the length of A is equal to 8cms, by using the Pythagorean Theorem.

Example 2 – Find out the perimeter of a rectangle whose length is 150 meters and the diagonal is 170 meters.

pythagorean theorem

In a rectangle, each angle values 90 degrees. Thus, PQR is right angled at Q.

PQ2 + QR2 = PR2
PQ2 + 1502 = 1702
PQ2 = 1702 – 1502
PQ2 = 320 * 20
PQ2 = 6400
PQ = 80

Thus, the perimeter of the rectangle PQRS = 2 * length + breadth
= 2 * (150 + 80)
= 2 * 230
= 460 m

Thus, these are the word problems of the concept of the Pythagorean example.

Example 3 – A ladder, which is 10 m long, rests along a building. The base of the ladder is 8m away from the building. How high from the ground is the top of the ladder?

theorem 2

Using the Pythagorean Theorem,

C^2 = A^2 + B^2
10^2 = 8^2 + B^2
100 = 64 + B^2

Subtracting 64 from both sides,

100 – 64 = 64 – 64 + B^2
36 = B^2

taking the square root of both sides,

6 = B

The top of the ladder is 6 m from the ground.