Question

A ${60}^{\circ }$ angle is constructed with arc radius as 4 cm. Find the area of the $△BOC$.

• A. $4\sqrt{3}c{m}^2$
• B. $\sqrt{3}c{m}^2$
• C. $12c{m}^2$
• D. $16c{m}^2$

Answer 1

The correct option is A $4\sqrt{3}c{m}^2$

Steps of construction of ${60}^{\circ }$ angle:
Step 1 : Draw the given ray OA and taking O as centre and radius = 4 cm, draw an arc which intersects OA, say at a point B.
Step 2 : Taking B as centre and with the same radius, draw an arc intersecting the previously drawn arc at a point C.
Step 3 : Draw a ray OD Passing through C then $\angle BOC$ is the required angle of 60°.

In $△OBC$
OB = OC [radii of same arc]
OB = BC [Arcs of equal radius]
Then OB = OC = BC = 4 cm. [Same arc with same radii]
$\therefore$ $△OBC$ is equilateral triangle.

Area of an equilteral triangle
= $\frac{\sqrt{3}}{4}\times sid{e}^2=\frac{\sqrt{3}}{4}\times 4^2$

= $4\sqrt{3}c{m}^2$