Question

The hypotenuse of a right triangle is 1m more than twce .the shortest side . if the third side is 7m more than the shortest side, find the side of a triangle.​

Answer 1

Answer:

Let the shortest side (BC) = x m Then, hypotenuse (AC) = (2x - 1) m and third side (AB) = (x + 1) m Now in right AABC, We have AC2 = AB2+ BC2 [By Pythagoras Theorem)

(2x - 1)² = (x + 1)² + (x)²

4x² + 1 - 4x = x² + 1 + 2x + x²

4x² + 1 - 4x = 2x² + 2x + 1 → 2x = 0 Hence, required sides of the triangle are AB = x + 1 = 4 m.

2x² - 6x = 0

2x(x 3) = 0

or x - 30

X = 0

or x = 3

Since x = 0 is not possible.

So, x = 3

BC = x = 3 m

and AC 2x - 1 = 5 m.