Three equal circles, each of radius …
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Sia ? 6 years, 6 months ago
Let A, B, C, be the centres of these circles. Joint AB, BC, CA. Radius of each circle = 6cm . AB = BC = CA = 12 cm .Then,ABC is an equilateral triangle.
Then, Area of ∆ABC,
{tex}=\frac{\sqrt3}{4}×(side)^2\\ =\frac{\sqrt 3}{4}×(12)^2\\ =\frac{\sqrt 3}{4}×(12)×(12)\\ =36×1.73=62.28{/tex}
Area of sector of angle 60° and radius 6 cm
{tex}=\frac{\theta}{360}×πr^2\\{/tex}
{tex} =\frac{60}{360}×3.14×6×6\ cm^2=18.84\ cm^2{/tex}
Required area = Area of ∆ABC- 3 ×area of one sector {tex}{/tex} {tex}{/tex}
{tex}{/tex}=62.28-3× 18.84
{tex}{/tex} =(62.28- 56.52) cm{tex}^2{/tex}
{tex}{/tex}=5.76 cm{tex}^2{/tex}
{tex}{/tex}Area of shaded region = 5.76 cm2
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