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°


Answer:

^@ 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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