In Fig. arcs are drawn by taking vertices A , | vedclass

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Question

since, $ABC$ is an equilateral triangle.

$	herefore$ $\angle A=\angle B=\angle C=60^{\circ}$

and $A B=B C=A C=10 \,cm$

So, $E, F$ and $D$ ar

Vedclass · Accepted Answer

$39.25$

Vedclass · Answer

since, $ABC$ is an equilateral triangle.

$	herefore$ $\angle A=\angle B=\angle C=60^{\circ}$

and $A B=B C=A C=10 \,cm$

So, $E, F$ and $D$ are mid-points of the sides.

$	herefore \quad A E=E C=C D=B D=B F=F A=5 \,cm$

Now, area of sector $CDE =\frac{	heta \pi^{2}}{360}=\frac{60 	imes 3.14}{360}(5)^{2}$

$=\frac{3.14 	imes 25}{6}=\frac{78.5}{6}=13.0833\, cm ^{2}$

$	herefore \quad$ Area of shaded region $=3$ (Area of sector $CDE$)

$=3 	imes 13.0833$

$=39.25 \,cm ^{2}$

In $Fig.$ arcs are drawn by taking vertices $A , B$ and $C$ of an equilateral triangle of side $10 \, cm$. 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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