Question

A round table cover has six equal design as shown in figure. If radius of the cover is $15cm$ and $ABCDEF$, is a regular hexagon then the total cost of making the shaded designs at the rate of $Rs10$ per $\frac{44-21\sqrt{3}}{84}c{m}^2$ is equal to

• A. $Rs1000$
• B. $Rs20,000$
• C. $Rs13,500$
• D. $Rs15,200$

Answer 1

The correct option is C $Rs13,500$

$r=15cm$

consider segment APB cord AB is a side of hexogon

each chord will substitute

$\begin{array}{c}=\frac{{360}^0}{6}\\ ={60}^0\\ in\Delta OAB\\ \angle OBA=\angle OBA\\ \angle AOB={60}^0\\ \angle O+\angle A+CB={180}^0\\ \angle O+\angle A+CA={180}^0\\ 2\angle A={180}^0-{60}^0\\ \angle A={120}^0\\ \angle A={120}^0\\ \angle A=\angle B=\angle O={60}^0\end{array}$

since, $\Delta AOB$ is a equilateral triangle.

$\begin{array}{c}Areaof\sec torOAPB\\ Area=\frac{60}{360}\\ =\frac{1}{60}\times \frac{22}{7}\times 15\times 15\\ =\frac{825}{7}cm\\ areaofdesign=areaofOAPB-areaof\Delta OB\\ =\frac{825}{7}-\frac{225}{4}\sqrt{3}\\ Areaof6designs=6\left(\frac{825}{7}-\frac{225}{4}\sqrt{3}\right)\\ \cos tof\frac{44-21\sqrt{3}}{84}c{m}^2=10\\ \cos tof1c{m}^2=10\times \frac{84}{44-21\sqrt{3}}\\ \cos tof6\left(\frac{825}{7}-\frac{225}{4}\times 1.7\right)=\frac{10\times 84}{44-21\sqrt{3}}\times 6\left[\frac{825}{7}-\frac{225}{4}\sqrt{3}\right]\\ =\frac{10\times 84\times 6}{44-21\sqrt{3}}\left(\frac{825}{7}-\frac{225}{4}\right)=13500\\ =660.82\left(117.80-97.42\right)\\ =13502\\ \left(13501.9681\right)\end{array}$