Question

# With vertices $$A,B$$ and $$C$$ of a triangle $$ABC$$ as centres, arcs are drawn with radius $$6\ cm$$ each in figure. If $$AB=20\ cm,\ BC=48\ cm$$ and $$CA=52\ cm$$, then find the area of the shaded region. (Use $$\pi=3.14$$)

Solution

## Given:-$$a = BC = 48 \; cm$$$$b = AC = 52 \; cm$$$$c = AB = 20 \; cm$$Semi-perimeter of $$\triangle{ABC} = \cfrac{a+b+c}{2} = \cfrac{52 + 48 + 20}{2} = 60 \; cm$$By using Heron's formula,Area of triangle, $$= \sqrt{s \left( s - a \right) \left( s - b \right) \left( s - c \right)}$$$$= \sqrt{60 \left( 60 - 48 \right) \left( 60 - 52 \right) \left( 60 - 20 \right)}$$$$= 480 \; {cm}^{2}$$Now,Area of sectors $$= \cfrac{\pi \theta}{360} {r}^{2}$$$$= \cfrac{\pi {r}^{2}}{360} \left( {\theta}_{1} + {\theta}_{2} + {\theta}_{2} \right)$$$$= \cfrac{3.14 \times 36}{360} \times \left( 180 \right) = 56.52 \; {cm}^{2}$$Therefore,Area of shaded region $$=$$ Area of triangle $$-$$ Area of sectors$$\Rightarrow$$ Area of shaded region $$= 480 - 56.52 = 423.48 \; {cm}^{2}$$Mathematics

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