Class IX Math NCERT Solution for Surface Areas and Volumes 13.9

EXERCISE 13.9 (OPTIONAL)
Q1.   A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the figure). The thickness of the plank is 5 cm every-where. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 poise per cm2 and the rate of painting is 10 poise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Sol: Here, Height = 110 cm
                 Depth = 25 cm
                 Breadth = 85 cm
                 Thickness of the plank = 5 cm
           Cost of polishing
                 Area to be polished = [(110 × 85)] + [(85 × 25) × 2] + [(110 × 25) × 2] +[(5 × 110) × 2] + [(75 × 5) × 4] cm2
                 = [9350] + [4250] + [5500] + [1100] + [1500] cm2 = 21700 cm2
           Cost of polishing at the rate 20 poise per cm2
                 
           Cost of painting
           ∵Area to be painted = [(75 × 20) × 6] + [(90 × 20) × 2] + [90 × 75] cm2
                 = [9000 + 3600 + 6750] cm2
                 = 19350 cm2
           ∴Cost of painting at the rate of 10 paise per cm2
                 
           Total Expenses
                 Total expenses = (Cost of polishing) + (Cost of painting)
                 = Rs. 4340 + Rs.1935
                 = Rs. 6275
           Thus, the total required expense = Rs. 6275.
Q2.   The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm' and black paint costs 5 paise per cm2.
Sol: For spherical part
           Surface area of a sphere = 4πr2
           
           Surface area of a wooden sphere to be painted silver = [1386 – 7.071] cm2
                 = 1378.93 cm2
           ⇒Surface area of 8 wood spheres to be painted = 8 × 1378.93 cm2
                 = 11031.44 cm2
           Now, the cost of silver painting at the rate of 25 paise per cm2
                 
           For cylindrical part
           Radius of the base of the cylindrical part (R) = 1.5 cm
           Height of the cylindrical part (h) = 7 cm
           ∴ Curved surface area of the cylindrical part (pillar) = 2πRh
                 
           ⇒ Total curve surface area of 8 pillars = 8 × 66 cm2 = 528 cm2
           ∴ Cost of painting black for 8 pillars at the rate of 5 paise per cm2
                 
           Total cost of painting
           Total cost of painting = [Cost of silver painting] + [Cost of black painting]
                 = Rs. 2757.86 + Rs. 26.40
                 = Rs. 2784.26
Q3.   The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Sol: Let the original diameter = d
           
           
           Per cent decrease in surface area
                 Decrease in curved surface area = [Original surface area] – [Decreased surface area]
           
           Thus, the required per cent decrease in curved surface area is 43.75%

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