Question

In the given figure, AB is a chord of length 6 cm drawn in a circle with centre 'O'. If OM is the perpendicular bisector of length $\sqrt{27}$ cm to chord AB then, find the area of sector of OAYB. (Use $\pi$=$\frac{22}{7}$)

• A. 19.65 ${cm}^2$
  
• B. 15.23 ${cm}^2$
  
• C. 12.36 ${cm}^2$
  
• D. 18.85 ${cm}^2$
  

Answer 1

The correct option is D 18.85 ${cm}^2$


As, OM is the perpendicular bisector of AB
so, AM = MB = $\frac{AB}{2}$ = $\frac{6}{2}$ = 3 cm
$\Rightarrow$ MB = 3 cm
Given: OM = $\sqrt{27}$ cm

Consider $△$OMB,
Use Pythagoras theorem,
${OB}^2$ = ${OM}^2$ + ${MB}^2$
$\Rightarrow$ ${OB}^2$ = ${\left(\sqrt{27}\right)}^2$ + $3^2$
$\Rightarrow$ ${OB}^2$ = 27 + 9 = 36
$\Rightarrow$ OB = $\sqrt{36}$ cm
$\Rightarrow$ OB = 6 cm (radius)

In $△$OMB,
sin ($\angle$MOB) = $\frac{MB}{OB}$
$\Rightarrow$ sin ($\angle$MOB) = $\frac{3}{6}$
$\Rightarrow$sin ($\angle$MOB) = $\frac{1}{2}$ [As, sin ${30}^{\circ }$ = $\frac{1}{2}$]
$\Rightarrow$sin ($\angle$MOB) = sin ${30}^{\circ }$
$\Rightarrow$ $\angle$MOB = ${30}^{\circ }$

so, $\angle$AOB = 2 $\times$$\angle$MOB
$\Rightarrow$$\angle$AOB = ${60}^{\circ }$

Area of sector = $\frac{x^{\circ }}{{360}^{\circ }}$ $\times$ $\pi$$\times$ $r^2$

Area of sector OAYB = $\frac{{60}^{\circ }}{{360}^{\circ }}$ $\times$ $\frac{22}{7}$$\times$ $6^2$
$\Rightarrow$ Area of sector OAYB = 18.85 ${cm}^2$