For this case we have that by definition, the area of a square is given by:
[tex]A = l ^ 2[/tex]
In this case, the square is 9 \ in side, then:
[tex]A = 9 ^ 2\\A = 81 \ in ^ 2[/tex]
On the other hand, the area of a triangle is given by:
[tex]A = \frac {1} {2} b * h[/tex]
Where:
b: It is the base of the triangle
h: It is the height
We have according to the figure that:
[tex]b = 9-5 = 4 \ in\\h = 9-4 = 5 \ in[/tex]
So:
[tex]A = \frac {1} {2} (4 * 5)\\A = \frac {20} {2}\\A = 10 \ in ^ 2[/tex]
Finally, the area of the figure is:
[tex]A_ {t} = 81 \ in ^ 2-10 \ in ^ 2\\A_ {t} = 71 \ in ^ 2[/tex]
Answer:
[tex]A_ {t} = 71 \ in ^ 2[/tex]
