The area of the square \(ABCD\) in the attached figure is 16 \(cm^2\), while \(O\) is the center of the square. The area of the shaded region is:
-
4 \(cm^2\)
-
8 \(cm^2\)
-
12 \(cm^2\)
-
2 \(cm^2\)
-
6 \(cm^2\)
Try to solve it before checking the answer.
4 \(cm^2\)
Let \(A\) be the area of the shaded region. We have that:
A = \text{area of } \triangle ABE – \text{area of } \triangle ABO
\]
If \(L\) is the side of the square, then \( L^2 = 16 \, cm^2 \); therefore, \( L = 4 \, cm \).
The base of these triangles is \( \overline{AB} = 4 \, cm \) and their heights measure 4 \(cm\) and 2 \(cm\), respectively. Then,
\begin{aligned}
A &= \frac{1}{2} (4) (4) – \frac{1}{2} (4) (2)
\\[.5em]
&= 8 – 4
\\[.5em]
&= \boldsymbol{ 4 \, cm^2}
\end{aligned}
\]
\begin{aligned}
A = \frac{1}{2} (4) (4) – \frac{1}{2} (4) (2)
= 8 – 4
= \boldsymbol{ 4 \, cm^2}
\end{aligned}
\]