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In the Given Figure, a is the Centre of the Circle. ∠ Abc = 45° and Ac = 7 √ 2 Cm. Find the Area of Segment Bxc. - Geometry

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Sum

In the given figure, A is the centre of the circle. \[\angle\]ABC = 45° and AC = 7 \[\sqrt{2}\]cm. Find the area of segment BXC.

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Solution

In ∆ABC,
AB = AC        (Radii of the circle)
∴ ∠ACB = ∠ABC = 45º        (Equal sides have equal angles opposite to them) 
Using angle sum property, we have
∠ACB + ∠ABC + ∠BAC = 180º
∴ 45º + 45º + ∠BAC = 180º
⇒ ∠BAC = 180º − 90º = 90º
Here,
Radius of the circle, r = \[7\sqrt{2}\]cm Measure of arc BXC, θ = 90º
∴ Area of segment BXC

\[= r^2 \left( \frac{\pi\theta}{360° } - \frac{\sin\theta}{2} \right)\]

\[ = \left( 7\sqrt{2} \right)^2 \left( \frac{22}{7} \times \frac{90°}{360°} - \frac{\sin90°}{2} \right)\]

\[ = 98 \times \frac{22}{7} \times \frac{1}{4} - 98 \times \frac{1}{2}\]

\[ = 77 - 49\] 

\[ = 28 {cm}^2\]

Thus, the area of the segment BXC is 28 cm2.

 
Concept: Areas of Sector and Segment of a Circle
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APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 7 Mensuration
Practice set 7.4 | Q 1 | Page 159
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