Question

The area of an equilateral triangle is 49√3 cm². Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle . Find the area of the triangle not included in the circles .

Answer 1

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Answer 2

Answer:-

• Refer to the Figure in the Attachment

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• Given:-

Area of $\triangle ABC$ = $\bf{49\sqrt{3}}$

$\theta = \bf{60 \degree}$

Let each side of the $\traingle$ be $\bf{x \: cm}$.

Area of Equilateral Triangle = $\bf({\frac{\sqrt{3}}{4}})\times (side)^2$

$\implies 49 \sqrt{3} = {\frac{\sqrt{3}}{4}} \times x^2$

$\implies x^2 = 49\sqrt{3} \times{\frac{4}{\sqrt{3}}}$

$\implies x^2 = 49 \times 4$

$\implies x^2 = 196$

$\implies x = \sqrt{196}$

$\implies \bf{x = 14 \: cm}$

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Given that:-

Radius of the circle half the length of the side of the $\triangle ABC$.

Radius of the circle = $\frac{1}{2}\times 14$

$\bf\implies{7 \: cm}$

Area of Sector = $(\dfrac{\theta}{360})\times \pi r^2$

Area of Sector = $(\dfrac{60}{360})\times {\dfrac{22}{7}}\times (7)^2$

$\implies \dfrac{1}{6}\times 22 \times 7$

$\implies \dfrac{154}{6}$

$\bf\implies{\dfrac{77}{3} \: cm^2}$

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Required Area = $49 \sqrt {3} - 3 \times \dfrac{77}{3}$

$\implies 49 \sqrt{3} - 77$

$\implies 49 \times 1.73 - 77 \: \: \: \: \: \: \: \: [\bf{\sqrt{3}= 1.732}]$

$\implies 84.77 - 77 \: cm^2$

$\bf\implies{7.77 \: cm^2}$

Therefore,

The area of the triangle$[\triangle]$] not included in the circles is $\bf{7.77 \: cm^2}$