Question
Find the area enclosed by the following figure:
✓
Answer
The given shape contains a triangle and a rectains.
For rectangle, $l = 13\ cm$ and $b = 4\ cm$
$\therefore$ Area of rectangle $=l \times 13 \times 4=52 \mathrm{~cm}^2$
For tringle, base $(b)=5 \mathrm{~cm}$ and height $(h)=(16-4) \mathrm{cm}=12 \mathrm{~cm}$
$\therefore$ Area of triangle $\frac{1}{2} \times \mathrm{b} \times \mathrm{h}=\frac{1}{2} \times 5 \times 12=30 \mathrm{~cm}^2$
$\therefore$ Total area enclosed by the shape $=(52+30) \mathrm{cm}^2=82 \mathrm{~cm}^2$
For rectangle, $l = 13\ cm$ and $b = 4\ cm$
$\therefore$ Area of rectangle $=l \times 13 \times 4=52 \mathrm{~cm}^2$
For tringle, base $(b)=5 \mathrm{~cm}$ and height $(h)=(16-4) \mathrm{cm}=12 \mathrm{~cm}$
$\therefore$ Area of triangle $\frac{1}{2} \times \mathrm{b} \times \mathrm{h}=\frac{1}{2} \times 5 \times 12=30 \mathrm{~cm}^2$
$\therefore$ Total area enclosed by the shape $=(52+30) \mathrm{cm}^2=82 \mathrm{~cm}^2$
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