# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
A wooden bookshelf has external dimension as follows: Height = 110 cm, Depth = 25 cm, breadth = 85 cm (see Fig. 13.31). The thickness of the plank is 5 cm 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:
We have given that
Height of bookshelf = 110 cm
Depth of bookshelf = 25 cm
and Breadth of bookshelf = 85 cm
The thickness of the plank = 5 cm
Total surface area of external face (Shading portion) = 2 (25 × 110) + 2 (85 × 25) + 110 × 85 + 4(5 × 85) + 2(90 × 5)
= 5500 + 4250 + 9350 + 1700 + 900
= 21,700 cm2
Total surface area of internal faces (without shading portion)
= 6(20 × 75) + 2(90 × 20) + 90 × 75
= 9000 + 3600 + 6750
= 19350 cm2
Now, cost of polishing the external faces at the rate of 20 paise/cm2 is = 21700 × 0.20 = Rs. 4340
and cost of painting the internal faces at tire rate of 10 paise/cm2 is, the bookself = 4340 + 1935 = Rs. 6275

Question 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, 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 cm2 and black paint costs 5 paise per cm2.

Solution:
The surface area of one wooden sphere = 4πr2
= $$4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2}$$
So, total surface area of all the 8.
Wooden sphere = $$8 \times 4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2}$$ = 11088 cm2
Now, some areas of the sphere can’t be silver printed due to the support.
Support area = $$8 \times \frac{22}{7} \times(1.5)^{2}$$ = 56.57 cm2
Required area for silver painting = 11088 – 56.57 = 11031.43 cm2
Cost of silver paint = 11031.43 × 0.25 = Rs. 2757.85
Now, support is to be painted in black colour.
So, curved surface of cylindrical support of:
= 8 × 2 × $$\frac {22}{7}$$ × 7 × 15
= 528 cm2
Cost to paint the support in black = 528 × 0.05 = Rs. 26.40
Total cost of paint = 2759.85 + 26.40 = Rs. 2784.25

Question 3.
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution:
Let the diameter of sphere = d

∴ Radius of sphere = $$\frac{d}{2}$$
∴ Curved surface area of sphere = 4πr2
= $$4 \times \pi \times \frac{d}{2} \times \frac{d}{2}$$
= πd2
In second case,
diameter of new sphere = d – 25% of d
= d – $$\frac{25}{100} \times d$$
= $$\frac{3 d}{4}$$
∴ Radius of new sphere = $$\frac{3 d}{8}$$

∴ Curved surface area of new sphere = 4πr2
= $$4 \times \pi \times \frac{3 d}{8} \times \frac{3 d}{8}$$
= $$\frac{9 \pi d^{2}}{16}$$
Therefore, surface area decrease in second case is = πd2 – $$\frac{9 \pi d^{2}}{16}$$
= $$\frac{16 \pi d^{2}-9 \pi d^{2}}{16}$$
= $$\frac{7 \pi d^{2}}{16}$$
So, Percent decreasing = $$\frac{7 \pi d^{2}}{\frac{16}{\pi d^{2}} \times 100}$$
= $$\frac{7 \pi d^{2}}{16 \pi d^{2}} \times 100$$
= 43.75%
So, the curved surface area decrease in the second case is 43.75%.

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