Question

ABCD
is a square 2 arcs of same radil APD and bpc
drawn as shown in figure If length of arc apd is
22cm. Find area of shaded region.
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Answer 1

$area \: of \: the \: shaded \: region = area \: of \: the \: larger \: sector - area \: of \: smaller \: sector$

$= \frac{1}{6}r {}^{2}\pi - \frac{1}{6}r {}^{2}\pi$

$= \frac{1}{6} \times \frac{22}{7} \times 21 \times 21 - \frac{1}{6} \times \frac{22}{7} \times 7 \times 7$

$= 231cm {}^{2} - 25.6cm {}^{2}$

$= 205.4cm {}^{2}$

Answer 2

area \: of \: the \: shaded \: region = area \: of \: the \: larger \: sector - area \: of \: smaller \: sectorareaoftheshadedregion=areaofthelargersector−areaofsmallersector

= \frac{1}{6}r {}^{2}\pi - \frac{1}{6}r {}^{2}\pi=61r2π−61r2π

= \frac{1}{6} \times \frac{22}{7} \times 21 \times 21 - \frac{1}{6} \times \frac{22}{7} \times 7 \times 7=61×722×21×21−61×722×7×7

= 231cm {}^{2} - 25.6cm {}^{2}=231cm2−25.6cm2

= 205.4cm {}^{2}=205.4cm2

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