With the vertices A , B and C of a triangle A | vedclass

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Question

Area of the sector with angle $A$

$=\frac{\angle A }{360^{\circ}} 	imes \pi r^{2}=\frac{\angle A }{360^{\circ}} 	imes \pi 	imes(5)^{2}\, cm ^{2}

Vedclass · Accepted Answer

$296.75$

Vedclass · Answer

Area of the sector with angle $A$

$=\frac{\angle A }{360^{\circ}} 	imes \pi r^{2}=\frac{\angle A }{360^{\circ}} 	imes \pi 	imes(5)^{2}\, cm ^{2}$

Area of the sector with angle $B$

$=\frac{\angle B }{360^{\circ}} 	imes \pi r^{2}=\frac{\angle B }{360^{\circ}} 	imes \pi 	imes(5)^{2} \,cm ^{2}$

and the area of the sector with angle $C =\frac{\angle C }{360^{\circ}} 	imes \pi 	imes(5)^{2} \,cm ^{2}$

Therefore, sum of the areas (in $cm ^{2}$ ) of the three sectors

$=\frac{\angle A }{360^{\circ}} 	imes \pi 	imes(5)^{2}+\frac{\angle B }{360^{\circ}} 	imes \pi 	imes(5)^{2}+\frac{\angle C }{360^{\circ}} 	imes \pi 	imes(5)^{2}$

$=\frac{\angle A +\angle B +\angle C }{360^{\circ}} 	imes 25 \pi$

$=\frac{180^{\circ}}{360^{\circ}} 	imes 25 \pi cm ^{2}\left(ight.$ Because $\left.\angle A +\angle B +\angle C =180^{\circ}ight)$

$=25 	imes \frac{\pi}{2}\, cm ^{2}=25 	imes 1.57 \,cm ^{2}=39.25\, cm ^{2}$

Now, to find area of $	riangle ABC$, we find

$s =\frac{a+b+c}{2}=\frac{48+50+14}{2}\, cm =56 \,cm$

By Heron's Formula,

$\operatorname{ar}( ABC )=\sqrt{ s ( s -a)( s -b)( s -c)}$

$=\sqrt{56 	imes 8 	imes 6 	imes 42}\, cm ^{2}$

$=336\, cm ^{2}$

So, area of the shaded region $=$ area of the $	riangle ABC -$ area of the three sectors

$=(336-39.25) \,cm ^{2}=296.75\, cm ^{2}$

With the vertices $A , B$ and $C$ of a triangle $ABC$ as centres, arcs are drawn with radii $5 \,cm$ each as shown in $Fig.$ If $AB =14 \,cm , BC =48 \,cm$ and $CA =50\, cm ,$ then find the area of the shaded region. (Use $\pi=3.14)$ (in $cm^2$)

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