Question

A circle having radius 4 cm contains a chord of length 4 cm and subtends an angle of 60 degrees. Find the area of the minor segment of the chord. [Take π = 22/7 and √3=1.73]

• A. $2c{m}^2$
• B. $1.46c{m}^2$
• C. $0.5c{m}^2$
• D. $3c{m}^2$

Answer 1

The correct option is B

$1.46c{m}^2$

Area of sector POQ
$=\frac{\theta }{360°}\times \pi r^2$

$=\frac{60°}{360°}\times \pi (4{)}^2=8.38c{m}^2$

In triangle OSQ which is right angled at S,
$O{Q}^2=S{Q}^2+O{S}^2$

$\Rightarrow 16=4+O{S}^2$
$OS=2\sqrt{3}$

Area of triangle POQ
$=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times 4\times 2\sqrt{3}$
$=6.92c{m}^2$

Now,
Area of segment PSQR = Area of sector POQ - Area of triangle POQ
$=8.38–6.92c{m}^2$
$=1.46c{m}^2$