Question

Question 6
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

Answer 1

Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
$\Rightarrow BD=\frac{AB}{2}$
Since, AD is the median of the triangle
$\therefore AO$ = Radius of the circle $=\frac{2}{3}AD$
$\Rightarrow \frac{2}{3}AD=32cm$
$\Rightarrow AD=48cm$
$In\Delta ADB$,

By Pythagoras theorem,
$A{B}^2=A{D}^2+B{D}^2\Rightarrow A{B}^2={48}^2+(\frac{AB}{2}{)}^2\Rightarrow A{B}^2=2304+\frac{A{B}^2}{4}\Rightarrow \frac{3}{4}(A{B}^2)=2304\Rightarrow A{B}^2=3072\Rightarrow AB=32\sqrt{3}cmAreaof\Delta ADB=\frac{\sqrt{3}}{4}\times (32\sqrt{3})c{m}^2=768\sqrt{3}c{m}^2$
$Areaofcircle=\pi R^2=\frac{22}{7}\times 32\times 32=\frac{22528}{7}c{m}^2$
Area of the design = Area of circle - Area of $\Delta ADB$
$=(\frac{22528}{7}-768\sqrt{3})c{m}^2$