Question

Question 13
A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per $c{m}^2.(Use\sqrt{3}=1.7)$

Answer 1

Number of equal designs = 6
Radius of round table cover = 28 cm
Cost of making design = $0.35perc{m}^2$
$\angle O=\frac{{360}^{\circ }}{6}={60}^{\circ }$
$\angle AOB$ is isosceles as two sides are equal. (Radius of the circle)
$\therefore \angle A=\angle B$
Sum of all angles of triangle = ${180}^{\circ }$
$\angle A+\angle B+\angle O={180}^{\circ }$
$\Rightarrow 2\angle A={180}^{\circ }-{60}^{\circ }$
$\Rightarrow \angle A=\frac{{120}^{\circ }}{2}$
$\Rightarrow \angle A={60}^{\circ }$
Triangle is equilateral as $\angle A=\angle B=\angle C={60}^{\circ }$
Area of equilateral $\Delta$
$=\frac{\sqrt{3}}{4}\times (side{)}^2$
Area of equilateral $\Delta AOB$
$=\frac{\sqrt{3}}{4}\times (OA{)}^2$
$=\frac{\sqrt{3}}{4}\times {28}^2=333.2c{m}^2$
Area of the sector making angle $\theta =\left(\frac{\theta }{{360}^{\circ }}\right)\times \pi r^2$
Area of sector $ACB=\left(\frac{{60}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$
$=\frac{1}{6}\times \frac{22}{7}\times 28\times 28=410.66c{m}^2$
Area of design
= Area of sector ACB - Area of equilateral $\Delta AOB$
$=410.66c{m}^2-333.2c{m}^2=77.46c{m}^2$
Area of 6 design = $6\times 77.46c{m}^2=464.76c{m}^2$
Total cost of making design
$=464.76c{m}^2\times 0.35perc{m}^2=₹162.66$