A photograph of Billiard/Snooker table has dimensions as 1/10 th of its actual size as shown in Fig. 9.71. The portion excluding six holes each of diameter 0.5 cm needs to be polished at rate of ₹200 per m². Find the cost of polishing.
Solution:
Given, a photograph of a billiard/snooker table has dimensions as 1/10 th of its actual size.
The portion excluding six holes each of diameter 0.5 cm needs to be polished.
The rate of polishing is ₹200 per m².
We have to find the total cost of polishing.
Original length = 25(10) = 250 cm
Original breadth = 10(10) = 100 cm
Area of table = length × breadth
= 250(100)
= 25000 cm²
Area of circle = πr²
Given, diameter = 0.5 cm
Radius = 0.5/2 = 0.25 cm
= (22/7)(0.25)²
= 22(0.0625)/7
= 0.1964 cm²
Area of 6 holes = 6(0.1964)
= 1.179 cm²
Area to be polished = area of table - area of 6 holes
= 25000 - 1.179
= 24998.82 cm²
We know, 1 m = 100 cm
Area to be polished in m² = 24998.82/10000
= 2.4999 m²
Cost of polishing 1 m² = ₹200
Cost of polishing table = 200(2.4999)
= ₹499.98
Therefore, the total cost of polishing is ₹499.98.
✦ Try This: A table cover of dimensions 3m 25cm×2m 30cm is spread on a table. If 30 cm of the table cover is hanging all around the table, find the area of the table cover, which is hanging outside the top of the table. Also, find the cost of polishing the table top at Rs.16 per square metre.
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 11
NCERT Exemplar Class 7 Maths Chapter 9 Problem 131
A photograph of Billiard/Snooker table has dimensions as 1/10 th of its actual size as shown in Fig. 9.71. The portion excluding six holes each of diameter 0.5 cm needs to be polished at rate of ₹200 per m². Find the cost of polishing.
Summary:
A photograph of Billiard/Snooker table has dimensions as 1/10 th of its actual size as shown in Fig. 9.71. The portion excluding six holes each of diameter 0.5 cm needs to be polished at a rate of ₹200 per m². The cost of polishing is ₹499.98
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