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 cm^{2} and the rate of painting is 10 paise per cm^{2}, 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 cm^{2}

Total surface area of internal faces (without shading portion)

= 6(20 × 75) + 2(90 × 20) + 90 × 75

= 9000 + 3600 + 6750

= 19350 cm^{2}

Now, cost of polishing the external faces at the rate of 20 paise/cm^{2} is = 21700 × 0.20 = Rs. 4340

and cost of painting the internal faces at tire rate of 10 paise/cm^{2} 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 cm^{2} and black paint costs 5 paise per cm^{2}.

Solution:

The surface area of one wooden sphere = 4πr^{2}

= \(4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2}\)

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 cm^{2}

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 cm^{2}

Required area for silver painting = 11088 – 56.57 = 11031.43 cm^{2}

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 cm^{2}

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 = 4πr^{2}

= \(4 \times \pi \times \frac{d}{2} \times \frac{d}{2}\)

= πd^{2}

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 = 4πr^{2}

= \(4 \times \pi \times \frac{3 d}{8} \times \frac{3 d}{8}\)

= \(\frac{9 \pi d^{2}}{16}\)

Therefore, surface area decrease in second case is = πd^{2} – \(\frac{9 \pi d^{2}}{16}\)

= \(\frac{16 \pi d^{2}-9 \pi d^{2}}{16}\)

= \(\frac{7 \pi d^{2}}{16}\)

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%.