Question
Find the area enclosed by the following figure:
✓
Answer
The given shape contains a rectangle and a triangle.
For rectangle, $\mathrm{I}=15 \mathrm{~cm}$ and $\mathrm{b}=3 \mathrm{~cm}$
$\therefore$ Area of rectangle $=1 \times \mathrm{b}=15 \times 3=45 \mathrm{~cm}^2$
Accoriding to the figuer, $\mathrm{BE}=\mathrm{AB}-\mathrm{AE}=15-10=5 \mathrm{~cm}$
For triangle, base $(b)=B E=5 \mathrm{~cm}$ and height $(h)=4 \mathrm{~cm}$
$\therefore$ Area of $\triangle \mathrm{BEG}=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}=\frac{1}{2} \times 5 \times 4=10 \mathrm{~cm}^2$
$\therefore$ Total area enclosed by the shape $=(45+10) \mathrm{cm}^2=55 \mathrm{~cm}^2$
For rectangle, $\mathrm{I}=15 \mathrm{~cm}$ and $\mathrm{b}=3 \mathrm{~cm}$
$\therefore$ Area of rectangle $=1 \times \mathrm{b}=15 \times 3=45 \mathrm{~cm}^2$
Accoriding to the figuer, $\mathrm{BE}=\mathrm{AB}-\mathrm{AE}=15-10=5 \mathrm{~cm}$
For triangle, base $(b)=B E=5 \mathrm{~cm}$ and height $(h)=4 \mathrm{~cm}$
$\therefore$ Area of $\triangle \mathrm{BEG}=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}=\frac{1}{2} \times 5 \times 4=10 \mathrm{~cm}^2$
$\therefore$ Total area enclosed by the shape $=(45+10) \mathrm{cm}^2=55 \mathrm{~cm}^2$
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