As shown in the diagram,triangle ABC is an eq | vedclass

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(B) $1$. The side of the equilateral triangle $	riangle ABC$ is $a = 70 \, cm$.$2$. The area of $	riangle ABC = \frac{\sqrt{3}}{4} 	imes a^2 = \fra

Vedclass · Accepted Answer

$1477.58$

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(B) $1$. The side of the equilateral triangle $	riangle ABC$ is $a = 70 \, cm$.$2$. The area of $	riangle ABC = \frac{\sqrt{3}}{4} 	imes a^2 = \frac{1.73}{4} 	imes 70^2 = 0.4325 	imes 4900 = 2119.25 \, cm^2$.$3$. $P$ and $R$ are midpoints of $AB$ and $AC$,so $AP = AR = \frac{70}{2} = 35 \, cm$. This is the radius $r$ of the sector $APR$.$4$. Since $	riangle ABC$ is equilateral,$\angle A = 60^\circ$.$5$. The area of the sector $APR = \frac{	heta}{360^\circ} 	imes \pi r^2 = \frac{60}{360} 	imes \frac{22}{7} 	imes 35 	imes 35 = \frac{1}{6} 	imes 22 	imes 5 	imes 35 = \frac{3850}{6} \approx 641.67 \, cm^2$.$6$. The shaded area is the area of $	riangle ABC$ minus the area of the sector $APR$.$7$. Shaded Area $= 2119.25 - 641.67 = 1477.58 \, cm^2$.

As shown in the diagram,$\triangle ABC$ is an equilateral triangle in which $BC = 70 \, cm$ and $P$ and $R$ are midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. $\widehat{PQR}$ is an arc of $\odot(A, AP)$. Find the area of the shaded region. $(\sqrt{3} = 1.73)$ (in $cm^2$)

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