Question

In a circular table cover of radius $32$ cm. A design is formed leaving an equilateral triangle $ABC$ is the middle as shown it. Find the area if the design.

Answer 1

Radius of the circle $r=32cm$
$\therefore$ Area of the circle $=\pi r^2$
$=\frac{22}{7}\times (32{)}^2$
$=\frac{22}{7}\times 1024$
$=3218.28sq.cm$
ii) $\angle BAC={60}^o\therefore \angle BOC={120}^o$
($\because$ Area at the centre is double than the angle in the circumference).
Now in $\Delta OBC,BC\perp OD$
$\angle BOD=\angle COD={60}^o$
Now in $△OBC,BC\perp OD$
$\angle BOD=\angle COD={60}^o$
In $\perp △BOD.\frac{OD}{OB}=\cos {60}^o$
$\frac{OD}{32}=\frac{1}{2}$
$\therefore OD=16cm$
$\angle BOD={60}^o$
$\therefore \frac{BD}{OB}=\sin {60}^o$
$\frac{BD}{32}=\frac{\sqrt{3}}{2}$
$\therefore BD=\frac{\sqrt{3}}{2}\times 32$
$BD=16\sqrt{3}cm$
$BC=BD+DC=BD+BD=2BD(\because BD=DC)$
$\therefore BC=2\times 16\sqrt{3}=32\sqrt{3}cm$
$\therefore △ABC$ is an quadrilateral triangle.
Each side of this triangle,
at $=2\times 16\sqrt{3}=32\sqrt{3}cm$
Area of quadrilateral triangle, $△ABC$
$=\frac{\sqrt{3}{a}^2}{4}$
$=\frac{1.73}{4}\times (32\sqrt{3}{)}^2$
$=\frac{1.73}{4}\times 1024\times 3$
$=1328.64sq.cm$
iii) Area of shaded region:
= Area of circle -Area of quadrilateral ${△}^{le}$
$=3218.28-1328.64$
$=1889.6sq.cm$