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.

Let XY = x m Therefore, YZ = (100 – x) m
In ∆ XYZ, ∠Z = 90° Therefore, by Pythagoras theorem XY^{2} = YZ^{2} + XZ^{2}⇒ x ^{2} = (100 – x)^{2} + 60^{2}⇒ ^{2}^{2} |

⇒ 200x = 10000 + 3600

⇒ 200x = 13600

⇒ x = 13600/200

⇒ x = 68

Therefore, distance between X and Y = 68 meters.

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

Given: h

So, we get

PR

h

⇒ 128 = 2k

⇒ 128/2 = k

⇒ 64 = k

⇒ √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

⇒ PS⇒ PS

⇒ PS

⇒ PS

⇒ PS

⇒ PS

⇒ 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,

x⇒ x

⇒ x

⇒ x

⇒ x

⇒ 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,

AD⇒ AD

⇒ AD

⇒ AD

⇒ 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.

**Conditions for the Congruence of Triangles**

**Right Angle Hypotenuse Side congruence**

**Converse of Pythagorean Theorem**

**7th Grade Math Problems**

**8th Grade Math Practice**

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