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 theoremXY2 = YZ2 + XZ2
⇒ x2 = (100 – x)2 + 602
⇒ 200x = 10000 + 3600
⇒ 200x = 13600
⇒ x = 13600/200
⇒ x = 68
Therefore, distance between X and Y = 68 meters.
⇒ √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.
In a rectangle, each angle measures 90°.
Therefore PSR is right angled at S
Using Pythagoras theorem, we get⇒ PS2 + SR2 = PR2
⇒ 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.
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
⇒ 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.
The vertical buildings AB and CD are 34 m and 29 m respectively.
Draw DE ┴ AB
Then AE = AB – EB but EB = BC
AE = 34 m - 29 m = 5 m
Now, AED is right angled triangle and right angled at E.
Therefor,AD2 = AE2 + ED2
⇒ 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.