Question

${\red{\underline{ \: \: Question:- }}}$
$\tt \: In \: figure, a \: square \: OABC \: is \\ \tt inscribed \: in \: a \: quadrant \: OPBQ. \\ \tt If \: OA=20 \: cm, \: find \: the \\ \tt area \: of \: the \: shaded \: region. \\ \ \textless \ br /\ \textgreater \ \tt(Use \: π=3.14)$
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Answer 1

${\underline{\underline{\purple{Solution.}}}}$

$\sf Each \: side \: of \: the \: square \: OABC$

$\: \: \: \: \: \textbf = \: 20 \: cm$

$\sf By \: Pythagoras \: Theorem$

$\: \: \: \: \rm \: OB {}^{2} = OA {}^{2} + (AB) {}^{2}$

$\: \: \: \: \: \: \: \: \: \: \rm = (20) {}^{2} + (20) {}^{2}$

$\: \: \: \: \: \: \: \: \rm = 400 + 400 = 800$

$\therefore \: \: \: \: \: \sf \: OB = \sqrt{800} = \sqrt{400 \times 2 \: }$

$\: \: \: \: \: \: \rm = 20 \sqrt{2 \ \: } cm$

$\sf \: i.e., \: \: \: \: \: \: \: radius \: of \: the \: circle = 20 \sqrt{2 \: } \: cm$

$\rm Area \: of \: the \: shaded \: region$

$\: \: \: \: \: \sf = Area \: of \: the \: quadrant \: OPBQ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - Area \: of \: the \: square \: OABC$

$\rm = \: \frac{90°}{360°}\pi(20 \sqrt{2) {}^{2} } - 20 \times 20$

$\: \: \: \: \: \rm = \: \: 200\pi - 400$

$\: \: \: \: \: \: \rm = \: \: \: 200 \times 3.14 - 400$

$\: \: \: \: \: \: \: \rm = \: \: \: \: 628 - 400$

$\: \: \: \: \: \: \: \: \: \: \: \sf \red{ = 228 \: cm {}^{2} }$

Answer 2

$\huge\underline\red{Given}$

OABC is a square

OA = 20cm

$\bold\green{NOTE:-}$

$\textbf{Each side of square is equal}$

$\textbf{Each angle of square is 90°}$

$\huge\underline\red{Formula}$$\huge\underline\red{used}$

(hypotenuse)²= (perpendicular)²+(base)²

Area of square = side×side

Area of sector = πr²×∅/360°

$\huge\underline\red{Solution}$

$\textbf{OABC is a square}$

$\textbf{By Pythagoras theorem}$

$\bold{(OB)^2\:=\:(OA)^2\:(AB)^2}$

$\bold{OB^2\:=(20)^2\:+\:(20)^2}$

$\bold{OB^2\:=\:400\:+\:400}$

OB = √800

$\bold{OB\:=\:20\:\sqrt{2}\:cm}$

So, radius of sector = 20√2cm

________________________________

$\textbf{Area of sector OPBQO}$

πr²×∅/360°

$\bold{3.14}$ × 20√2 × 20√2 × $\large\frac{90}{360}$

3.14 × 800 × $\large\frac{1}{4}$

3.14 × 200

628cm²

________________________________

$\textbf{Area of square OABC}$

= side × side

= 20 × 20

= 400cm²

_______________________________

$\textbf{Area of shaded region}$

= 628 - 400

= 228cm²