 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 71 of Exercise 10E

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71
##### The hypotenuse of a right – angled triangle is 1 metre less than twice the shortest side. If the third side is 1 metre more than the shortest side, find the sides of the triangle.

Let the shortest side of triangle be xm

According to the question ;

Hypotenuse = 2x – 1 m

Third side = x + 1 m

Applying Pythagoras theorem

(2x – 1)2 = (x + 1)2 + x2

4x2 – 4x + 1 = x2 + 2x + 1 + x2 using (a – b)2 = a2 – 2ab + b2

2 x2 – 6x = 0

2x (x – 3) = 0

x = 0 or x = 3

Length of side cannot be 0 thus the shortest side is 3m

Hypotenuse = 2x – 1 = 6 – 1 = 5m

Third side = x + 1 = 3 + 1 = 4m

Thus the dimensions of triangle are 3m, 4m and 5m.

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