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43 votes
The figure above shows a big square that contains 2 smaller squares in orange and four half circles in red. If the side of the big containing square is 9 inches long, what is the total area of orange?A. 40.16 in2B. 41.62 in2C. 42.53 in2D. 43.49 in2

The figure above shows a big square that contains 2 smaller squares in orange and-example-1
User NSGaga
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1 Answer

11 votes
11 votes

To solve this question we will use the following diagram:

Therefore:


\begin{gathered} 2r_1+r_1=9in, \\ 2r_2+r_2=2r_(1.) \end{gathered}

Adding like terms in the above equations we get:


\begin{gathered} 3r_1=9in, \\ 3r_2=2r_1\text{.} \end{gathered}

Therefore:


\begin{gathered} r_1=(9in)/(3)=3in, \\ r_2=(6in)/(3)=2in\text{.} \end{gathered}

Now, notice that the orange region is formed by 2 semicircles of radius 3in and 2 semicircles of radius 1in, then, the area is:


\begin{gathered} A=\pi(3in)^2+\pi(2in)^2 \\ =9\pi in^2+4\pi in^2 \\ =13\pi in^2\approx40.84in^2. \end{gathered}

Answer: 40.84in².

The figure above shows a big square that contains 2 smaller squares in orange and-example-1
User Steve Cohen
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