Question

What is the area of the figure below ireadya 68in.b 43in.c 88in.d 48in.

Answer 1

We can divide the figure in these two:

So, let's find first the area of the square and the trapeze:

`\begin{gathered} A_s=l^2 \\ \Rightarrow A_s=(3)^2=9in^2 \\ A_t=(B+b)/(2)\cdot h \\ \Rightarrow A_t=(7+3)/(2)\cdot5=(10)/(2)\cdot5=5\cdot5=25 \\ A_t=25in^2 \end{gathered}`

Now, notice that the figure is two trapezes and two squares, then the total area will be two times the area of the square plus two times the area of the trapeze. We get the following:

`\begin{gathered} A=2\cdot A_s+2\cdot A_t \\ \Rightarrow A=2\cdot9+2\cdot25=18+50=68 \\ A=68in^2 \end{gathered}`

therefore, the total area of the figure is 68 in^2