Question

In figure 7.43, A is the centre of the circle. ∠ABC=45° and AC=7√2 cm. Find the area of segment BXC.

Answer 1

Steps:
1)
Radius of circle, r =7√2 cm
Since,
/ABC = /ACB = 45° (Angles opposite. to equal sides are equal)
By Angle sum property of triangle,
/ABC = 180° -45° -45° = 90°

Angle of sector,theta = 90°


2) We know,
Area of segment ,
$= \frac{ {r}^{2} }{2} ( \frac{\pi \theta}{180 \degree} - \sin( \theta) ) \\ = > \frac{( {7 \sqrt{2} )}^{2} }{2} ( \frac{\pi \times 90\degree}{180 \degree} - \sin( 90 \degree) ) \\ = > 49( \frac{\pi}{2} - 1) \: {cm}^{2}$

Hence, Area of segment is
$\boxed { 49( \frac{\pi}{2} - 1) {cm}^{2} }$

Answer 2

Answer:

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