Question

A circumcircle is drawn around an equilateral triangle. If the circum-radius r = 14 cm, find the area of the shaded region.

• A. 61.68 cm²
• B. 65.32 cm²
• C. 69.51 cm²
• D. 64.23 cm²

Answer 1

The correct option is A

61.68 cm²

The shaded region can be divided into three segments.

$\delta$ ABC is an equilateral triangle.
Hence all the angles are equal = 60°
Consider chord AB, the angle made by the chord at circumference = $\angle ACB$ = 60°

Now, let us consider the chord AB.

Chord AB makes an angle of 60° at C, i.e. $\angle ACB={60}^{\circ }$

Then, the angle $\angle AOB=2\angle ACB$

= $2\times {60}^{\circ }$ ( angle a chord makes at the centre is double the angle the chord makes at any point on the circumference of the circle)

= ${120}^{\circ }$

Now let us consider the sector AOB,

The area of the sector AOB = $\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

= $\frac{120}{360}\times \frac{22}{7}\times 7^2$

= $\frac{1}{3}\times 22\times 14\times 2$

= $\frac{616}{3}$

= 105.33$c{m}^2$

Now that we know the area of the sector, we need to find the area of triangle AOB in order to find the area of the segment ABD.

Draw OE ⊥ AB ;

AE = $\frac{1}{2}$ AB ( Perpendicular to a chord passing through the centre will bisects the chord

Here $\angle AOE=\frac{1}{2}\angle AOB$

= $\frac{1}{2}\times 120$

= ${60}^{\circ }$

By applying trigonometric identities in rt. triangle AOE we get,

AE = OA sin 60

= $\frac{\sqrt{3}r}{2}$

= $14\times \frac{\sqrt{3}}{2}$

= $\sqrt{3}$ cm

EO = OA Cos 60

=$r\times \frac{1}{2}$

= $\frac{14}{2}$

= 7 cm

In triangle AOB, AB = 2 AE (OE is the perpendicular bisector of AB)

Therefore, AB = 2 x 7 $\sqrt{3}$

= 14 $\sqrt{3}$ cm

Area of the triangle AOB = $\frac{1}{2}$ x AB x OE

= $\frac{1}{2}\times 14\sqrt{3}\times 7$

= $7\sqrt{3}\times 7$

= $49\sqrt{3}$

= 49 x 1.73

= 84.77 $c{m}^2$

Therefore, the area of the segment ABD = Area of the sector AOBD – Area of the triangle AOB

= 105.33 – 84.77 = 20.56 $c{m}^2$

The area of 1 segment = 20.56$c{m}^2$

Therefore the area of the total segment = 3 x area of 1 segment = 3 x 20.56 = 61.68 $c{m}^2$

Therefore the total area of the shaded region is 61.68 $c{m}^2$