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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.
NCERT Solutions for Class 9 Maths Chapter 13 - Exercise 13.9 Surface Areas and VolumesSolution:
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 cmarea = 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.
NCERT Solutions for Class 9 Maths Chapter 13 - Exercise 13.9 Surface Areas and VolumesSolution:
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)= 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 r= (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
= NCERT Solutions for Class 9 Maths Chapter 13 - Exercise 13.9 Surface Areas and Volumes
= (7d2 / 16d2) × 100 = 700 / 16 = 43.75%.
Therefore, the percentage decrease in the surface area of the sphere is 43.75%.

The document NCERT Solutions for Class 9 Maths Chapter 13 - Exercise 13.9 Surface Areas and Volumes is a part of the Bank Exams Course NCERT Mathematics for Competitive Exams.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 13 - Exercise 13.9 Surface Areas and Volumes

1. What are the different types of surfaces and volumes covered in NCERT Solutions for Exercise 13.9?
Ans. Exercise 13.9 of NCERT Solutions for Surface Areas and Volumes covers the calculation of the surface area and volume of various geometrical shapes such as cubes, cuboids, cylinders, cones, and spheres.
2. How can I calculate the surface area of a cube mentioned in Exercise 13.9 of NCERT Solutions?
Ans. To calculate the surface area of a cube, you need to multiply the length of any side of the cube by itself and then multiply the result by 6. The formula for the surface area of a cube is SA = 6a^2, where "a" represents the length of any side of the cube.
3. How do I find the volume of a cylinder mentioned in Exercise 13.9 of NCERT Solutions?
Ans. The volume of a cylinder can be calculated by multiplying the area of the base (which is the circle) by the height of the cylinder. The formula for the volume of a cylinder is V = πr^2h, where "r" represents the radius of the base and "h" represents the height of the cylinder.
4. Can you provide an example of finding the surface area of a cone mentioned in Exercise 13.9 of NCERT Solutions?
Ans. Sure! Let's say the radius of the base of a cone is 5 cm and the slant height is 12 cm. To find the surface area, you first need to calculate the curved surface area by using the formula CSA = πrl, where "r" represents the radius and "l" represents the slant height. Then, you calculate the total surface area by adding the curved surface area to the area of the base. The formula for the total surface area of a cone is TSA = CSA + πr^2.
5. How can I determine the volume of a sphere mentioned in Exercise 13.9 of NCERT Solutions?
Ans. The volume of a sphere can be found by using the formula V = (4/3)πr^3, where "r" represents the radius of the sphere. To calculate the volume, simply substitute the value of the radius into the formula and perform the necessary calculations.
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