For what value of x, the triangle having the sides x, x+7, x+8 is a right angled triangle

x = 5

Step-by-step explanation:

Using Pythagoras' identity.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

Here the longest side, the hypotenuse is x + 8, thus

x² + (x + 7) ² = (x + 8) ² ← expand factors using FOIL

x² + x² + 14x + 49 = x² + 16x + 64

2x² + 14x + 49 = x² + 16x + 64 ← subtract terms on right side from both sides

x² - 2x - 15 = 0 ← in standard form

(x - 5) (x + 3) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 3 = 0 ⇒ x = - 3

However x > 0, thus x = 5

The sides of the right triangle are 5, 12, 13