Question

with vertices A,B,C of a triangle ABC as centres,arcs are drawn with radii 6cm each.if AB=20cm,BC=48cm,CA=52cm find the area of shadedbregion

Answer 1

Ans. Hope this answers helps u !

Answer 2

Answer:

$\red {Area \:of \:the \: shaded \: region}$

$\green {= 423.52\:cm^{2}}$

Step-by-step explanation:

$Given \: A,B\: and \: C \:are \: vertices \: of \\\triangle ABC . Arcs \: are \: drawn \: with \: radii \\6\:cm \:each .\\AB = c = 20\:cm ,\:BC = a = 48\:cm , \: CA = b = 52 \:cm .$

$\underline { Finding \: \triangle ABC\: area }$

$\blue { (By \: Heron's \: Formula) }$

$s = \frac{a+b+c}{2} \\= \frac{48 + 52 + 20 }{2}\\= \frac{ 120}{2} = 60$

$\triangle = \sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{60(60-48)(60-52)(60-20)}\\=\sqrt{60\times 12\times 8 \times 40}\\= \sqrt { 480\times 480}\\= 480\:cm^{2}\:---(1)$

$Area \: of \: 3 \: sectors = \frac{(\angle A+\angle B + \angle C )}{360}\times \pi r^{2}$

$= \frac{180}{360} \times 3.14 \times 6^{2}$

$= 3.14 \times 18\\= 56.52 \:cm^{2}\:--(2)$

$Area \:of \:the \: shaded \: region \\= 480 - 56.52\\= 423.52 \:cm^{2}$

Therefore.,

$\red {Area \:of \:the \: shaded \: region}$

$\green {= 423.52\:cm^{2}}$

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