Question

A ${60}^{\circ }$ angle is constructed with arc radius as 3 cm. Find the perimeter of the $△BOC$ in cm.

• A. 9

Answer 1

The correct option is A 9
Steps of construction of ${60}^{\circ }$ angle:
Step 1 : Draw the given ray OA and taking O as centre and radius = 3 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 = 3 cm. [Same arc with same radii]
$\therefore$ $△OBC$ is equilateral triangle.

Perimeter of equilateral triangle
= OB + OC + BC
= 3 + 3 + 3 = 9 cm.