Question

In the given figure, ABC is the right-angled triangle at B. A semicircle is drawn on AB as the diameter. If AB=12cm and BC=5cm, then the area of the shaded region is?​

Answer 1

Answer:

Correct option is A)

In △ABC,∠A=90

o

By Pythagoras theorem,

BC

2

=AB

2

+AC

2

BC

2

=3

2

+4

2

9+16=25

∴BC=5cm

A(shadded region)=A(△ ABC)+A (semicircle AB) + (semicircle AC) - A(semicircle BAC)

=

2

1

×3×4+

2

1

×π(

2

3

)

2

+

2

1

×π(

2

4

)

2

−

2

1

×π(

2

5

)

2

=6+

8

9π

+2π−

8

25π

=6+2π−2π

=6cm

2

Answer 2

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

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$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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In the given figure, ABC is a right angled triangle right angle at B , AB is diameter =12 cm , BC =5 cm

Area of shaded portion = area of semicircle + area of right $\rm \triangle ABC$

$\[ \begin{array}{l} \\ \displaystyle\rm =\frac{1}{2} \pi r^{2}+\frac{1}{2} \times AB \times BC \\\\ \displaystyle\rm =\frac{1}{2} \pi(6)^{2}+\frac{1}{2} \times 12 \times 5 cm ^{2} \\\\ \boxed{ \color{darkcyan}\displaystyle\rm =(18 \pi+30) cm ^{2} }\end{array} \]$

Hence, Correct option is (b)