Question

The three sides of a right angled triangle are x,x+3 and x+6. Find
the value of x. Also find the area of right angled triangle.​

Answer 1

Answer:

x= 9 or x= -3

Step-by-step explanation:

Pythagoras theorem,

x^2 + (x+3)^2 = (x+6)^2

x^2 + x^2+9+6x=x^2+36+12x

2x^2-x^2+6x-12x+9-36=0

x^2-6x-27=0

x^2-9x+3x-27=0

x(x-9) +3(x-9)=0

(x-9)(x+3)=0

x=9 or x=-3

Answer 2

Answer :-

In a right angled triangle,

$\sf P^2 + B^2 = H^2$

As the hypotenuse is the longest side,

Here, hypotenuse = x + 6

➩ $\sf x^2 + (x+3)^2 = (x+6)^2$

➩ $\sf x^2 + x^2 + 9 + 6x = x^2 + 36 + 12x$

➩ $\sf 2x^2 + 6x + 9 = x^2 + 36 + 12x$

➩ $\sf x^2 + 6x - 12x + 9 - 36 = 0$

➩ $\sf x^2 - 6x - 27 = 0$

➩ $\sf x^2 + 3x - 9x - 27 = 0$

➩ $\sf x ( x + 3 ) - 9 ( x + 3 ) = 0$

➩ $\sf (x-9)(x+3) = 0$

➩ $\sf x = 9,-3$

As negative side is not possible,

$\boxed{\sf x = 9}$

Area of right angled triangle = 1/2 × B × P

$\sf = 1/2 \times 9 \times ( 9 + 3 )$

$\sf = 1/2 \times 9 \times 12$

$\sf = 54$

$\boxed{\sf Area = 54}$