Question

find the area of semicicle in the shaded region and take pie=3.14 and for finding AC use Pythagoras theorem ​

Answer 1

Answer:

area of shaded region = area of semicircle - area of triangle

ar(shaded) = πr²- 1/2 base × height

= (3.14) - 1/2 (base* height

by pythagorous theprem

12²+16²=AC²

144+256= Ac²

Ac= √400.

AC= 20. base=20

then height =r = 10

ar(shaded)= 3.14(10)² - 1/2(20*10)

ar(shaded) = 3.14×100 - 1/2×200

ar(shaded)= 314 - 100

=214sq.cm

therefore area of shaded region is 214sq.cm

Answer 2

☞︎︎︎ Given :

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• Base = 16cm

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• height = 12cm

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• Hint for finding AC

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• Pie = 3.14

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☞︎︎︎ To Find :

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• Area if shaded region

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$\looparrowright$ we see that, whole figure is of semi circle

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☞︎︎︎ Solution :

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• ☕︎ Pythagoras Theorem -

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$\large \dashrightarrow \boxed{\boxed{ \mathtt{ {h}^{2} = {b}^{2} + {p}^{2} }}}$

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❥︎ here,

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• h = hypotenuse (side to find )

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• b = 16cm

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• p = 12cm

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✰ On solving :

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$\mapsto \: \mathtt{ {h}^{2} = {16}^{2} + {12}^{2} }$

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✞︎ We know ,

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• 16² = 256

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• 12² = 144

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$\mapsto \mathtt{ \: h {}^{2} = 256 + 144 }$

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$\mapsto \mathtt{ \: {h}^{2} = 400 }$

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$\looparrowright$ On transposing the terms :

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$\mapsto \: \mathtt{h = \sqrt{400} }$

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$\bigstar \: \large \boxed{ \therefore \mathtt{ \: h = 20cm}}$

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➪ Now,

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• Find area of triangle

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$\large \dashrightarrow \boxed{ \mathtt{area \: = \frac{1}{2} \times b \times h}}$

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✞︎ Here,

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• b = base (16cm)

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• h = height (12cm)

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✰ On Solving :

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$\mapsto \mathtt{ \: area = \dfrac{1}{2} \times 16 \times 12}$

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$\implies \mathtt{ \: area = 8 \times 12}$

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$\bigstar \: \large \boxed{ \therefore \mathtt{ \underline{ \: area = 96 {cm}^{2} }}}$

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➪ Now ,

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☞︎︎︎ We Have :

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• Area of triangle

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• hypotenuse

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➪ Finding area of semicircle -

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• pie = 3.14

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• r = radius (20cm)

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$\large \dashrightarrow \boxed{ \mathtt{area \: = \frac{1}{2} \times \pi{r}^{2} }}$

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❥︎ We take half because it is a semicircle

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$\implies \mathtt{ \: area \: = \dfrac{1}{2} \times 3.14 \times 20 {}^{2} }$

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$\implies \mathtt{ \: area \: = \dfrac{1}{2} \times 3.14 \times 20 {}^{} \times 20 }$

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$\implies \mathtt{ \: area \: = {1}{} \times 3.14 \times 10 {}^{} \times 20 }$

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$\implies \mathtt{ \: area \: = 3.14 \times 200 }$

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☕︎ Break the decimal :

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$\implies \mathtt{ \: area \: = \dfrac{314}{100} \times 200 }$

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$\implies \mathtt{ \: area \: = 314 \times 2 }$

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$\bigstar \: \large \boxed{ \therefore \underline{ {\mathtt{ \: area \: = 628 {cm2} }}}}$

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☞︎︎︎ Area of shaded region :

= Area of semicircle - Area of triangle

$☞︎︎︎ \mathtt{ \: \: 628 {cm}^{2} - 96 {cm}^{2} }$

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$\: \large \bigstar \: \boxed{ \boxed{\therefore { \mathtt{ \: area \: = 532 {cm {}^{2} } }}}} \bigstar$

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