Question

A round table cover has six equal designs as shown in figure. If the radius of the cover is $28$cm, find the cost of making the design at the rate of Rs.$3.50$ per $c{m}^2$. (Use $\sqrt{3}=1.7$).

• A. Rs. $1145.3$
• B. Rs. $1348$
• C. Rs. $1626.8$
• D. Rs. $1723.5$

Answer 1

Answer: (C)

Rs. $1626.8$

$r=28cm,\theta =\frac{{360}^0}{6}={60}^0$

Area $OAMB=\frac{\theta }{{360}^0}\times \pi r^2=\frac{{60}^0}{{360}^0}\times \frac{22}{7}\times 28\times 28$

$=$ $\frac{1232}{3}{cm}^2=410.67{cm}^2$

Now, in $\Delta ONA$ and $\Delta ONB$

i) $\angle ONA=\angle ONB\left[each{90}^0\right]$

ii) $OA=OB$ [Radii of common circle]

iii) $ON=ON$ [common]

$\therefore \Delta ONA\cong \Delta ONB\left[ByRHScongruency\right]$

Hence, $AN=NB=\frac{1}{2}AB$

& $\angle AON=\angle BON=\frac{1}{2}\angle AOB=\frac{{60}^0}{2}={30}^0$

Now in $\Delta ONA,{\cos 30}^0=\frac{ON}{OA}\Rightarrow \frac{\sqrt{3}}{2}=\frac{ON}{28}\Rightarrow ON=14\sqrt{3}cm$

& ${\sin 30}^0=\frac{AN}{OA}\Rightarrow \frac{1}{2}=\frac{AN}{28}\Rightarrow AN=14cm$

& $2AN=14\times 2=28cm=AB$

$\therefore$ $ar\Delta AOB=\frac{1}{2}\times AB\times ON=\frac{1}{2}\times 28\times 14\sqrt{3}=196\sqrt{3}=196\times 1.7$ $=333.2{cm}^2$

$\therefore$ Area of one design $=410.67-333.2=77.47$ ${cm}^2$

$\therefore$ Area of six design= $6\times 77.47=464.82{cm}^2$

Therefore, Cost of making the designs$=Rs.(464.82$$\times$3.5$)

$=Rs.1626.8$