Given alongside is a geometrical shape. C\ is centre of semicircle with radius BC=21cm. \triangle ABCis isosceles triangle. (Use \pi=3.14)

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Solution

Q.
Given alongside is a geometrical shape.
C is centre of semicircle with radius BC=21 cm.
$△ A B C$ is isosceles triangle. $(U s e π = 3.14)$

Which of the following statements are correct

$(i) A r e a o f t r i a n g l e A B C o u t s i d e c i r c u l a r r e g i o n = 47.4075 c m ²$

$(i i) A r e a o f s e c t o r C B D = 276 c m ²$

$(i i i) A n g l e o f s e c t o r C B D = 45 °$

$(i v) A r e a o f ∆ A B C = 220.5 c m ²$

$C h o o s e t h e c o r r e c t o p t i o n f r o m t h e f o l l o w i n g$

1 (i), (iii) and (iv)

2 (i) and (ii)

3 (i) and (iii)

4 (iii) and (iv)

Ans.
(1)

$S i n c e ∆ A B C i s i s o s c e l e s A B = B C = 21 cm Area of ∆ ABC : = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times (21)^{2} = 220.5 {cm}^{2} ∴ Statement (iv) is correct . ∆ ABC is isosceles with ∠ B = 90 ° So, the other angles each measure 45 ° . \Rightarrow Angle of sector CBD = 45 ° ∴ Statement (iii) is correct . Area of sector CBD = \frac{θ}{360 °} \times {πr}^{2} = \frac{45}{360} \times 3.14 \times 21 \times 21 = 173.0925 {cm}^{2} ∴ Statement (ii) is incorrect . Area of triangle outside circular region = Area of ∆ ABC - Area of sector CBD = 220.5 - 173.0925 = 47.4075 {cm}^{2} ∴ Statement (i) is correct .$

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