Question

Question 131

A photograph of BIlliard/ Snooker table has dimensions as $\frac{1}{10}th$ of its actual size as shown in the given figure.

The portion excluding six holes each of diameter 0.5 cm needs to be polished at the rate of ₹ 200 per $m^2$. Find the cost of polishing.

Answer 1

Actual length = $25\times 10=250cm$

Actual breadth = $10\times 10=100cm$

Area of table = $250\times 100=25000c{m}^2$

Actual radius of 1 hole $=\frac{0.5\times 10}{2}=2.5cm$

Area of 6 holes $=6\times \pi r^2=6\times \frac{22}{7}\times 2.5\times 2.5=117.85c{m}^2$

Area of portion excluding holes = 25000 - 117.85 = 24882.15 $c{m}^2$

$\therefore$ Cost of polishing
$=₹\frac{24882.15}{10000}\times 200=₹497.64=₹500(approx.)$