Question

Find the difference of the areas of two segments of a circle formed by a chord of length $5cm$. Subtending an angle of ${90}^o$ till the centre.

• A. $32.14$ sq. cm
• B. $35.42$ sq. cm
• C. $38.96$ sq. cm
• D. $42.43$ sq. cm

Answer 1

Answer: (A)

$32.14$ sq. cm

Let $r$ be the radius of circle and $AB$ be the chord, which makes ${90}^o$ angle at centre.

$AB=5cm$

In right $△OAB$, using pythagoras theorem

$O{A}^2+O{B}^2=A{B}^2\Rightarrow r^2+r^2=5^2\Rightarrow 2r^2=25\Rightarrow r=\frac{5}{\sqrt{2}}$

Area of circle $=\pi r^2=\frac{22}{7}\times \frac{5}{\sqrt{2}}\times \frac{5}{\sqrt{2}}=39.28c{m}^2$

Area of minor segment $=Areaofsector-Areaof△OAB=\frac{{90}^o}{{360}^o}\times \pi r^2-\frac{1}{2}\times \frac{5}{\sqrt{2}}\times \frac{5}{\sqrt{2}}=\frac{1}{4}\times 39.28-\frac{25}{4}=\frac{14.28}{4}=3.57c{m}^2$

Area of major segment $=39.28-3.57=35.71c{m}^2$

Required difference $=35.71-3.57=32.14c{m}^2$