Question

The figure shown is made up of triangle ABF and rectangle BCDE. The area of the figure is 90 square feet. If AB=BC=BF=FE, find AC

Answer 1

The complete question in the attached figure

let
x-----> distance AB
so
AC=2x

we know that
area of the figure=area of triangle+area of rectangle
area of the figure=90 ft²

step 1
 find the area of triangle
$A=b*h/2$
b=x
h=x
so
$A=(x*x)/2$
Area of triangle=x²/2 ft²

step 2
find the area of rectangle
$A=b*h$
b=2x
h=x
$A=(2x)*x------\ \textgreater \ A=2x^{2}$
area rectangle=2x²

step 3 
find value of x
area of the figure=(x²/2)+2x²
area of the figure=90 ft²
90=(x²/2)+2x²-----> multiply by 2 both sides----> 180=x²+4x²
5x²=180-----> x²=36-----> x=6 ft

AC=2x-----> AC=2*6----> AC=12 ft

the answer is
AC=12 ft