### A chord of a circle of radius $42 \space cm$ subtends a right angle at the center. Find the area of the corresponding major segment. $\left[ \pi = \dfrac { 22 } { 7 } \right]$ O A B 90°

$5040 \space cm^2$

Step by Step Explanation:
1. We know that area of minor sector $= \dfrac { \theta } { 360 } \times \pi r^2$

Substituting the value of $\theta$ and $r$ in the formula, we have

Area of sector $OAB = \dfrac { 90 } { 360 } \times \dfrac { 22 } { 7 } \times (42)^2 = 1386 \space cm^2$
2. Also, area of a right-angled triangle = $\dfrac { 1 } { 2 } \times Base \times Height$

So, area of right-angled $\triangle OAB = \dfrac { 1 } { 2 } \times OA \times OB = \dfrac { 1 } { 2 } \times 42 \times 42 = 882 \space cm^2$
3. Now, area of minor segment = area of minor sector $OAB$ $-$ area of $\triangle OAB = 1386 - 882 = 504 \space cm^2$
4. Area of major segment $=$ area of the circle $-$ area of minor segment $= \pi (42)^2 - 504 \\ = 5544 - 504 \\ = 5040 \space cm^2$
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