Word problems on Pythagorean Theorem

Learn how to solve different types of word problems on Pythagorean Theorem.

Pythagoras Theorem can be used to solve the problems step-by-step when we know the length of two sides of a right angled triangle and we need to get the length of the third side.

Three cases of word problems on Pythagorean Theorem:

Case 1: To find the hypotenuse where perpendicular and base are given.

Case 2: To find the base where perpendicular and hypotenuse are given.

Case 3: To find the perpendicular where base and hypotenuse are given.

Word problems using the Pythagorean Theorem:

1. A person has to walk 100 m to go from position X in the north of east direction to the position B and then to the west of Y to reach finally at position Z. The position Z is situated at the north of X and at a distance of 60 m from X. Find the distance between X and Y.

⇒ 200x = 10000 + 3600

⇒ 200x = 13600

⇒ x = 13600/200

⇒ x = 68

Therefore, distance between X and Y = 68 meters.

2. If the square of the hypotenuse of an isosceles right triangle is 128 cm², find the length of each side.

Solution:

Let the two equal side of right angled isosceles triangle, right angled at Q be k cm.

Given: h² = 128

So, we get

PR² = PQ² + QR²

h² = k² + k²

⇒ 128 = 2k²

⇒ 128/2 = k²

⇒ 64 = k²

⇒ √64 = k

⇒ 8 = k

Therefore, length of each side is 8 cm.

Using the formula solve more word problems on Pythagorean Theorem.

3. Find the perimeter of a rectangle whose length is 150 m and the diagonal is 170 m.

Solution:

In a rectangle, each angle measures 90°.

Therefore PSR is right angled at S

Using Pythagoras theorem, we get

⇒ PS² + SR² = PR²

⇒ PS² + 150² = 170²

⇒ PS² = 170² – 150²

⇒ PS²= (170 + 150) (170 – 150), [using the formula of a² - b² = (a + b) (a - b)]

⇒ PS²= 320 × 20

⇒ PS² = 6400

⇒ PS = √6400

⇒ PS = 80

Therefore perimeter of the rectangle PQRS = 2 (length + width)

                                                          = 2 (150 + 80) m

                                                          = 2 (230) m

                                                          = 460 m

4. A ladder 13 m long is placed on the ground in such a way that it touches the top of a vertical wall 12 m high. Find the distance of the foot of the ladder from the bottom of the wall.

Solution:

Let the required distance be x meters. Here, the ladder, the wall and the ground from a right-angled triangle. The ladder is the hypotenuse of that triangle.

According to Pythagorean Theorem,

x² + 12² = 13²

⇒ x² = 13² – 12²

⇒ x² = (13 + 12) (13 – 12)

⇒ x² = (25) (1)

⇒ x² = 25

⇒ x = √25

⇒ x = 5

Therefore, distance of the foot of the ladder from the bottom of the wall = 5 meters.

5. The height of two building is 34 m and 29 m respectively. If the distance between the two building is 12 m, find the distance between their tops.

Solution:

The vertical buildings AB and CD are 34 m and 29 m respectively.

Draw DE ┴ AB

Then AE = AB – EB but EB = BC

Therefore AE = 34 m - 29 m = 5 m

Now, AED is right angled triangle and right angled at E.

Therefor,

AD² = AE² + ED²

⇒ AD² = 5² + 12²

⇒ AD² = 25 + 144

⇒ AD² = 169

⇒ AD = √169

⇒ AD = 13

Therefore the distance between their tops = 13 m.

The examples will help us to solve various types of word problems on Pythagorean Theorem.

Congruent Shapes

Congruent Line-segments

Congruent Angles

Congruent Triangles

Conditions for the Congruence of Triangles

Side Side Side Congruence

Side Angle Side Congruence

Angle Side Angle Congruence

Angle Angle Side Congruence

Right Angle Hypotenuse Side congruence

Pythagorean Theorem

Proof of Pythagorean Theorem

Converse of Pythagorean Theorem

7th Grade Math Problems

8th Grade Math Practice

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