The hypotenuse of a right triangle is 6m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the area(in m²) of the triangle.

Correct option is C

Given:
Hypotenuse = 6 + 2x(shortest side)
Second side = hypotenuse -2
Formula Used:
Pythagoras’ theorem
Area of triangle = $$\frac12× b× h$$​​
Solution:
Let smallest side be x
Then hypotenuse = 2x+6
Second side = 2x+6-2= 2x+4
Using Pythagoras’ theorem
​$$H^2 = P^2 + B^2\\(2x + 6)^2 = x^2 + (2x + 4)^2\\4x^2 + 36 + 24x = x^2 + 4x^2 + 16 + 16x\\x^2 - 8x - 20 = 0\\x = 10, \, x = -2\\$$​​
Other Side = 2x+4 = 24m
Area of triangle = $$\frac{1}{2} b \times h = \frac{1}{2} \times 10 \times 24 = 120 \, \text{m}^2$$​​