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(B) The sum of the areas of the three sectors at vertices $A, B,$ and $C$ is given by:Area $= \frac{\angle A}{360^{\circ}} 	imes \pi r^2 + \frac{\ang

Vedclass · Accepted Answer

$296.75$

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(B) The sum of the areas of the three sectors at vertices $A, B,$ and $C$ is given by:Area $= \frac{\angle A}{360^{\circ}} 	imes \pi r^2 + \frac{\angle B}{360^{\circ}} 	imes \pi r^2 + \frac{\angle C}{360^{\circ}} 	imes \pi r^2$$= \frac{(\angle A + \angle B + \angle C)}{360^{\circ}} 	imes \pi r^2$Since the sum of angles in a triangle is $180^{\circ}$,the area of the three sectors is:$= \frac{180^{\circ}}{360^{\circ}} 	imes 3.14 	imes (5)^2 = 0.5 	imes 3.14 	imes 25 = 39.25 \, cm^2$Now,we calculate the area of $	riangle ABC$. Since $14^2 + 48^2 = 196 + 2304 = 2500 = 50^2$,the triangle is a right-angled triangle with base $48 \, cm$ and height $14 \, cm$.Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 48 	imes 14 = 336 \, cm^2$The area of the shaded region is the area of the triangle minus the area of the three sectors:Area $= 336 - 39.25 = 296.75 \, cm^2$

With the vertices $A, B$ and $C$ of a triangle $ABC$ as centres,arcs are drawn with radii $5 \, cm$ each as shown in the figure. 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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