NCERT Solutions for Class 9 Maths Chapter 13 - Exercise 13.9 Surface Areas and Volumes

Q.1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm, Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.
Solution: External dimensions of book self,
Length, l = 85cm
Breadth, b = 25 cm
Height, h = 110 cm
External surface area of shelf while leaving out the front face of the shelf
= lh + 2(lb + bh)
= [85 × 110 + 2(85 × 25 + 25 × 110)] = (9350 + 9750) = 19100
External surface area of shelf is 19100 cm²
Area of front face = [85 × 110 - 75 × 100 + 2(75 × 5)] = 1850 + 750
So, area is 2600 cm²
Area to be polished = (19100 + 2600) cm² = 21700 cm².
Cost of polishing 1 cm² area = Rs 0.20
Cost of polishing 21700 cm² area Rs. (21700 × 0.20) = Rs 4340
Dimensions of row of the book shelf
Length(l) = 75 cm
Breadth (b) = 20 cm and
Height(h) = 30 cm
Area to be painted in one row= 2(l + h)b + lh = [2(75 + 30) × 20 + 75 × 30] = (4200 + 2250) = 6450
So, area is 6450 cm².
Area to be painted in 3 rows = (3 × 6450)cm² = 19350 cm².
Cost of painting 1 cm² area = Rs. 0.10
Cost of painting 19350 cm² area = Rs (19350 x 0.1) = Rs 1935
Total expense required for polishing and painting = Rs. (4340 + 1935) = Rs. 6275
The cost for polishing and painting the surface of the book shelf is Rs. 6275.

Q.2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. Eight such spheres are used forth is 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 25 paise per cm² and black paint costs 5 paise per cm².
Solution: Diameter of wooden sphere = 21 cm
Radius of wooden sphere, r = diameter / 2 = (21 / 2) cm = 10.5 cm
Formula: Surface area of wooden sphere = 4πr²
= 4 × (22 / 7) × (10.5)² = 1386
So, surface area is 1386 cm³
Radius of the circular end of cylindrical support = 1.5 cm
Height of cylindrical support = 7 cm
Curved surface area = 2πrh
= 2 × (22 / 7) × 1.5 × 7 = 66
So, CSA is 66 cm²
Now, Area of the circular end of cylindrical support = πr²
= (22 / 7) × 1.52
= 7.07
Area of the circular end is 7.07 cm²
Again,
Area to be painted silver = [8 × (1386 - 7.07)] = 8 × 1378.93 = 11031.44
Area to be painted is 11031.44 cm²
Cost for painting with silver colour = Rs(11031.44 × 0.25) =Rs 2757.86
Area to be painted black = (8 × 66) cm² = 528 cm²
Cost for painting with black colour = Rs (528 × 0.05) = Rs26.40
Therefore, the total painting cost is: = Rs(2757.86 + 26.40) = Rs 2784.26.

Q.3. The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution: Let the diameter of the sphere be “d”.
Radius of sphere, r₁ = d / 2
New radius of sphere, say r₂ = (d / 2) × (1 - 25 / 100) = 3d / 8
Curved surface area of sphere, (CSA)₁ = 4πr₁² = 4π × (d / 2)² = πd² …(1)
Curved surface area of sphere when radius is decreased (CSA)² = 4πr₂² = 4π × (3d / 8)² = (9 / 16)πd² …(2)
From equation (1) and (2), we have
Decrease in surface area of sphere = (CSA)₁ – (CSA)₂
= πd² – (9 / 16)πd²
= (7/16)πd².
Percentage decrease in surface area of sphere
=
= (7d² / 16d²) × 100 = 700 / 16 = 43.75%.
Therefore, the percentage decrease in the surface area of the sphere is 43.75%.