A triangular-shaped wall has a base of 2x + 4 and a height of x + 3. The area of the triangle is 56 in.2. What is the value of x?

Area of the triangle :
(1/2) (2x + 4) (x + 3) = 56
(x + 2) (x + 3) = 56
x² + 3x + 2x + 6 = 56
x² + 5x - 50 = 0
(x + 10)(x - 5) = 0
x = -10 (rejected) or x = 5

Ans: x = 5

(x = -10 is rejected for the height = -10 + 3 = -7)

The area of a triangle is calculated by A = 1/2(B)(H). From the values you provided, B = 2x+4 and H = x+3. The value of x can be found by writing an equation and solving.

1/2(2x+4)(x+3) = 56

2*(1/2) (2x+4)(x+3) = 56(2)

(x+3)(2x+4) = 112

2x^2+10x+12 = 112

2x^2 +10x -100 =0

The hard part here is solving it once you get it into quadratic form, i.e. ax^2+bx+c = 0.

In this case: 2x^2 +10x - 100 = 0

I solved it by graphing, but you can use the quadratic formula.

-10±√(10^2 -4* 2(-100)) / 2*2

x=5, -10 (ignore -10 bc it is extraneous)

To check our work we plug everything back in...

(2*5 + 4) = 14 (base)
(5+3) = 8 (height)

A = 1/2(14)(8) = 56

I hope this helped!

(x+2)(x+3)=56
solve for x , with x>0