In the rectangle $ABDF$ above, $C$ and $E$ are midpoints of sides $\overline{BD}$ and $\overline{AB}$, respectively. What fraction of the area of the rectangle is shaded?

เฉลยละเอียด

[STEP]Assume the length of $\overline{BD}$ and $\overline{AB}$[/STEP]

Assume that$$

\overline{BC}=\overline{CD}=1

$$and$$

\overline{AE}=\overline{EB}=1

$$so that$$

\overline{AB}=\overline{BD}=2

$$

[STEP]Calculate the triangles $EBC$ and $CDF$'s areas[/STEP]

Therefore the area of the rectangular $ABDF$ is $2\times2=4$ square unit. Then we calculate the area of the triangle $EBC$ and $CDF$. After that to get the shaded area, we substract the triangles areas from the rectangular area.

\begin{eqnarray*}
 \text{Area of }EBC &=& \frac12\times 1\times 1\\
&=&\frac12\\
 \text{Area of }CDF &=& \frac12\times1\times2 \\
&=& 1
\end{eqnarray*}

[STEP]Calculate the shaded area and the ratio[/STEP]

So the shaded area is

$$4-\frac12 - 1 = \frac52$$

and finally the fraction of the shaded part is

$$\frac{\frac52}{4} = \frac58$$ 

[ANS]$\frac58$[/ANS]