Question

❛ Find the areas of the shaded region in the given figure..❜

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Answer 1

$\large\underline{\bold{Solution :- }}$

It is given that

• Radius of big semi-circle, R = 14 cm.

So,

• Area of big semi-circle is given by

$\rm :\longmapsto\:A_1 = \: \dfrac{1}{2} \: \pi \: {R}^{2}$

$\rm :\longmapsto\:A_1 = \dfrac{1}{2} \times \dfrac{22}{7} \times {14}^{2}$

$\rm :\longmapsto\: \boxed{ \bf \: A_1 \: = \: 308 \: {cm}^{2} }$

Now,

• Radius of small semi-circle, r = 7 cm

So,

• Area of 2 small semi-circle are given by

$\rm :\longmapsto\:A_2 = \:2 \times \: \dfrac{1}{2} \: \pi \: {r}^{2}$

$\rm :\longmapsto\:A_2 = \: \pi \: {r}^{2}$

$\rm :\longmapsto\:A_2 = \: \dfrac{22}{7} \times {7}^{2}$

$\rm :\implies\: \boxed{ \bf \: A_2 \: = \: 154 \: {cm}^{2} }$

Hence,

• Total area of shaded region is given by

$\rm :\longmapsto\:\tt \: Area_{(shaded \: region)} = A_1 + A_2$

$\rm :\longmapsto\:\tt \: Area_{(shaded \: region)} = 308 + 154$

$\rm :\longmapsto\: \boxed{\tt \: Area_{(shaded \: region)} = 462 \: {cm}^{2}}$

Additional Information :-

$\rm :\longmapsto\:\tt \: Area_{(circle)} = \pi \: {r}^{2}$

$\rm :\longmapsto\:\tt \: Area_{(sector)} = \dfrac{\pi \: {r}^{2} \theta}{360}$

$\rm :\longmapsto\:\tt \: Area_{(sector)} = \dfrac{1}{2} lr$

$\rm :\longmapsto\:\tt \: Area_{(major \: sector)} = \dfrac{\pi \: {r}^{2} (360 - \theta)}{360}$

$\rm :\longmapsto\:\tt \: Area_{(minor \: segment)} = \dfrac{ {r}^{2} }{2} \bigg(\dfrac{\pi \:\theta}{180} - sin\theta \bigg)$

$\rm :\longmapsto\:\tt \: Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4} {(side)}^{2}$