Question

ABCP is a quadrant of a circle of radius 14cm. with AC as diameter, a semi-circle is drawn. Find the area of the shaded region. ( take π=22/7)​

Answer 1

Answer:

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Answer 2

GIVEN:-

$\bf \: ABCPA \: is \: a \: quadrant \: of \: a \: circle \: of \: radius=14cm$

TO SOLVE:-

$\bf The \: area \: of \: shaded \: reason=?$

SOLUTION:-

$\bf \: We \: have,r=14cm$

$\bf \therefore AB=BC=14cm \: and \angle ABC=90⁰$

$\implies \bf \: AC= \sqrt{AB²+BC²} \: \: [By \: pythagorus \: theorum]$

$\bf \implies \sqrt{(14)²+(14)²} cm$

$\bf \implies \sqrt{2×196} cm$

$\bf \implies 14 \sqrt{2}cm$

$\bf \therefore \: Radius \: of \: the \: circle= \frac{1}{2} ×AC = 7 \sqrt{2}cm$

$\bf \underline{Required \: area}=(area \: of \: the \: semicircle \: with \: AC \: as \: diameter)$

$\bf \: \: \: \: \: \: + (area \: of \triangle \: ABC)-(area \: of \:quadrant \: with \: r=14cm)$

$\bf \large[ \small \frac{1}{2} × \frac{22}{7} ×(7 \sqrt{2} )+ \frac{1}{2}×14×14- \frac{1}{4}× \frac{22}{7}×14×14 \large]$

$\bf \ = (154 + 98 - 154)cm {}^{2} = 98cm {}^{2}$

$\boxed{ \therefore \bf Required \: area=98cm²}$