In the figure,arcs are drawn by taking vertic | vedclass

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(D) Given that $ABC$ is an equilateral triangle with side length $10 \, cm$.Therefore,$\angle A = \angle B = \angle C = 60^{\circ}$.Since $D, E$ and $

Vedclass · Accepted Answer

$39.25$

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(D) Given that $ABC$ is an equilateral triangle with side length $10 \, cm$.Therefore,$\angle A = \angle B = \angle C = 60^{\circ}$.Since $D, E$ and $F$ are mid-points of sides $BC, CA$ and $AB$ respectively,the radius of each arc is $r = 5 \, cm$.The shaded region consists of three sectors,each with a central angle of $60^{\circ}$ and radius $r = 5 \, cm$.Area of one sector $= \frac{	heta}{360^{\circ}} 	imes \pi r^2 = \frac{60^{\circ}}{360^{circ}} 	imes 3.14 	imes (5)^2$.Area of one sector $= \frac{1}{6} 	imes 3.14 	imes 25 = \frac{78.5}{6} \approx 13.0833 \, cm^2$.Total area of the shaded region $= 3 	imes (	ext{Area of one sector}) = 3 	imes \frac{78.5}{6} = \frac{78.5}{2} = 39.25 \, cm^2$.

Part $I$	Part $II$
$1.$ Formula to find the length of a minor arc	$a.$ $C=2\pi r$
$2.$ Formula to find the area of a minor sector	$b.$ $A=\pi r^{2}$
$3.$ Formula to find the area of a circle	$c.$ $l=\frac{\pi r \theta}{180}$
$4.$ Formula to find the circumference of a circle	$d.$ $A=\frac{\pi r^{2} \theta}{360}$

In the figure,arcs are drawn by taking vertices $A, B$ and $C$ of an equilateral triangle of side $10 \, cm$ as centers to intersect the sides $BC, CA$ and $AB$ at their respective mid-points $D, E$ and $F$. Find the area of the shaded region (Use $\pi = 3.14$) (in $cm^2$).

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