Question

What is the area of the figure
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__in2

Answer 1

This is the area of a triangle and a square
Area of a triangle is 1/2BH
here total height is 16 in  triangle h is 16-10=6
base is 10
A=1/2*10*6
A=30 inch^2

Area of square=a^2
A2=10^2=100inches^2
A1+A2=100+30
130inches^2

Answer 2

Answer:

$A_{total}=112in^{2}$

Step-by-step explanation:

The given figure is formed by a right rectangle and a square.

By given, the dimensions of the square is 10 inches by 10 inches..

So, to find the dimensions of the right triangle, we just need to subtract.

The height of the right triangle would be

$h=16in-10in=6in$

Because the total length combined is 16 inches, and the side of the square is 10 inches. So, by subtraction, we find the height of the right triangle.

Applying the same reasoning, we find the base of the triangle

$b=10in-6in=4in$

Now, we are able to find both areas.

$A_{triangle}=\frac{bh}{2}=\frac{4(6)}{2}=12in^{2}$

$A_{square}=(10in)^{2}=100in^{2}$

Therefore, the whole are is the sum of each figure

$A_{total}=A_{triangle} +A_{square} =12in^{2} +100in^{2} =112in^{2}$