Question

The front compound wall of a house is decorated by wooden spheres of diameter $21cm$, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius $1.5cm$ and height $7cm$ and is to be painted black. Find the cost of paint required if silver paint costs $25paiseperc{m}^2$ and black paint costs $5paiseperc{m}^2$.

Answer 1

Diameter of each sphere $=21cm$

$\therefore$ Radius $\left(R\right)=\frac{21}{2}cm$

Radius of each cylinder $\left(r\right)=1.5cm$ and height $\left(h\right)=7cm$

Now surface area of one sphere$=4\pi R^2$

$=4\times \frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}=1386c{m}^2$ and surface area of one cylinder $=2\pi rh$

$=2\times \frac{22}{7}\times 1.5\times 7=66c{m}^2$

$\therefore$ Surface area of $8$ spheres $=8\times 1386=11088c{m}^2$ and surface area of $8$ cylinders $=8\times 66=528c{m}^2$

Now, surface area of $8$ cylinders tops$=8\times \pi r^2$

$=8\times \frac{22}{7}\times 1.5\times 1.5=56.57c{m}^2$

$\therefore$ Surface area of spheres excluding base area $=11088-56.57=11031.43c{m}^2$

As, rate of silver painting the spheres $=25pperc{m}^2$

$\therefore$ Cost of silver painting $11088c{m}^2$

$=Rs\frac{11031.43\times 25}{100}=Rs2757.86$

Rate of black painting the cylinder $=5pperc{m}^2$

$\therefore$ Cost of black painting $528c{m}^2$

$=Rs\frac{528\times \times 5}{100}=\frac{2640}{100}=Rs26.40$

Hence, total cost $=2757.86+26.40=Rs2784.26$.