Question

A chord of circle of radius \[10\;cm\] subtends a right angle at the centre. Find the area of the major sector.

Answer 1

Hint: Firstly check the angle subtended by the chords to the major sector of the circle. Then find the area of that part out of the total area of the circle using the formulae of area of sector.

Complete step-by-step answer:
A chord of circle of radius \[10\;cm\] subtends a right angle at the centre.

As shown in the diagram, the major sector of the circle is $ OAB $ and the angle subtended by the chord to the center is $ 90^\circ $ . So, the angle subtended in the minor sector is equal to $ 90^\circ $ .
The total angle is always equal to $ 360^\circ $ . So, the angle subtended by the chord in the major sector is equal to $ 360^\circ - 90^\circ = 270^\circ $ .
The formula for the area of the sector of circle with angle $ \theta $ with the chord and radius $ r $ is equal to $ \dfrac{\theta }{{360^\circ }} \times \pi {r^2} $ .
Substitute the radius and angle in the formula for the area of the major sector:
 $
\Rightarrow A = \dfrac{\theta }{{360^\circ }} \times \pi {r^2} \\
   = \dfrac{{270^\circ }}{{360^\circ }} \times \dfrac{{22}}{7} \times 10\;cm \times 10\;cm \\
   = \dfrac{3}{4} \times 314 \\
   = 235.5\;c{m^2} \;
  $
So, the area of the major sector of the circle is equal to $ 235.5\;c{m^2} $ .
So, the correct answer is “$ 235.5\;c{m^2} $ ”.

Note: The right angle subtended by the chord of the circle at the center lies in the minor sector of the circle and the total angle is always equal to $ 360^\circ $ . The formula for the area of the sector of circle with angle $ \theta $ with the chord and radius $ r $ is equal to $ \dfrac{\theta }{{360^\circ }} \times \pi {r^2} $ .