Question
In Fig below, $ABCD$ is a rectangle with dimensions $32m$ by $18m$. $ADE$ is a triangle such that $\text{EF}\bot\text{AD}$ and $EF = 14\ cm$. Calculate the area of the shaded region.
✓
Answer
We have,
$\text { Area of the rectangle }=A B \times B C=32 \mathrm{~m} \times 18 \mathrm{~m}=576 \mathrm{~m}^2$
$\text { Area of the triangle }=\frac{1}{2}(\mathrm{AD} \times \mathrm{FE})$
$=\frac{1}{2}(\mathrm{BC} \times \mathrm{FE})[\text { Since } A D=B C]$
$=\frac{1}{2}(18 \mathrm{~m} \times 14 \mathrm{~m})$
$=9 \mathrm{~m} \times 14 \mathrm{~m}$
$=126 \mathrm{~m}^2$
Area of the shaded region $=$ Area of the rectangle - Area of the triangle
$=(576-126) \mathrm{m}^2$
$=450 \mathrm{~m}^2$
$\text { Area of the rectangle }=A B \times B C=32 \mathrm{~m} \times 18 \mathrm{~m}=576 \mathrm{~m}^2$
$\text { Area of the triangle }=\frac{1}{2}(\mathrm{AD} \times \mathrm{FE})$
$=\frac{1}{2}(\mathrm{BC} \times \mathrm{FE})[\text { Since } A D=B C]$
$=\frac{1}{2}(18 \mathrm{~m} \times 14 \mathrm{~m})$
$=9 \mathrm{~m} \times 14 \mathrm{~m}$
$=126 \mathrm{~m}^2$
Area of the shaded region $=$ Area of the rectangle - Area of the triangle
$=(576-126) \mathrm{m}^2$
$=450 \mathrm{~m}^2$
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