Find the diameter of the circle whose area is | vedclass

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Question

Here, radius $r_{1}$ of first circle $=\frac{20}{2} \,cm =10 \,cm$ and radius $r_{2}$ of the second circle $=\frac{48}{2}\, cm =24 \,cm$

Therefore,

Vedclass · Accepted Answer

$52$

Vedclass · Answer

Here, radius $r_{1}$ of first circle $=\frac{20}{2} \,cm =10 \,cm$ and radius $r_{2}$ of the second circle $=\frac{48}{2}\, cm =24 \,cm$

Therefore, sum of their areas $=\pi r_{1}^{2}+\pi r_{2}^{2}=\pi(10)^{2}+\pi(24)^{2}=\pi 	imes 676$ ..........$(1)$

Let the radus of the new curcle be $r cm$. Its area $=\pi r^{2}$ ..........$(2)$

Therefore, from $(1)$ and $(2),$

$\pi r^{2}=\pi 	imes 676$

$r^{2}=676$

$r=26$

Thus, radius of the new circle $=26\, cm$

Hence, diameter of the new circle $=2 	imes 26 \, cm =52 \,cm$

Part $I$	Part $II$
$1.$ $\overline{ OA } \cup \overline{ OB } \cup \widehat{ APB }$	$a.$ Major sector
$2.$ $\overline{ AB } \cup \widehat{ AQB }$	$b.$ Minor segment
$3.$ $\overline{ AB } \cup \widehat{ APB }$	$c.$ Minor sector
$4.$ $\overline{ OA } \cup \overline{ OB } \cup \widehat{ AQB }$	$d.$ Major segment

Find the diameter of the circle whose area is equal to the sum of the areas of the two circles of diameters $20\, cm$ and $48 \,cm .$ (in $cm$)

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