Question

In a circular table cover of radius 12 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks



Each side of the equilateral triangle subtends ${120}^{\circ }$ at the centre.

The area of the each sector $=\frac{120}{360}\times \pi \times {12}^2=150.72c{m}^2$

Perpendicular from chord is drawn to the centre. By RHS congruence, the two triangles will be congruent.

The perpendicular divides the chord into equal halves.

The angle subtended by each triangle at the centre is ${60}^{\circ }$.

Height of perpendicular $=rcos{60}^{\circ }=12\times 0.5=6cm.$

Length of chord $=2\times r\times sin{60}^{\circ }=24\times \frac{\sqrt{(}3)}{2}=20.6cm$

The area of each small triangle $=0.5\times 20.6\times 6=61.8c{m}^2$

The area of each segment $=150.72-61.8=88.92c{m}^2$

The total area is $=88.92\times 3=266.76c{m}^2$

The area of circle $=\pi \times {12}^2=452.16c{m}^2$