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

NCERT Solutions for Class 9 Maths Chapter 13 Surface
Areas and Volumes Ex 13.9 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 13 Surface
Areas and Volumes Ex 13.9.

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

**Ex 13.9 Class 9 Maths Question 1.****
**A wooden bookshelf has external
dimensions as follows:

Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). 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:**

Here, length (l) = 85 cm, breadth (b) = 25 cm and height (h) = 110 cm

External surface area = Area of four faces + Area of back + Area of front beading

= [2 (110 + 85) × 25 + 110 × 85 + (110 × 5 × 2) + (75 × 5) × 4] cm

^{2}= 21700 cm

^{2}

∴ Cost of polishing external faces = Rs. (21700 × 20/10) = Rs. 4340

Internal surface area = Area of five faces of 3 cuboids each of dimensions 75 cm × 30 cm × 20 cm

=
Total surface area of 3 cuboids of dimensions 75 cm × 30 cm × 20 cm – Area of
bases of 3 cuboids of dimensions 75 cm ×
30 cm × 20 cm

=
3(2(75 × 30 + 30 × 20 + 75 × 20)) cm^{2} –
3 × (75 ×
30) cm^{2}

= 6(2250 + 600 + 1500) cm^{2} – 6750 cm^{2} = 19350
cm^{2}

∴ Cost of painting inner faces = Rs. 19350 × 10/100 = Rs.
1935

Here, total expenses = Rs. (4340 + 1935) = Rs. 6275

**Ex
13.9 Class 9 Maths 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 figure.
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:
**Here, diameter of a sphere = 21 cm

Radius of a sphere (r) = 21/2 cm

Surface area of a sphere = 4Ï€r

^{2}

∴ Surface area of 8 spheres

= 8 × 4 × 22/7 × (21/2)^{2
}cm^{2}

**Ex 13.9 Class 9 Maths Question 3.****
**The diameter of a sphere is
decreased by 25%. By what per cent does its curved surface area decrease?

**Solution:****
**Let the diameter of the sphere be d.

After decreasing, diameter of the sphere

= d – 25/100 × d

= d – ¼ d = ¾ d

Since, surface area of a sphere = 4Ï€r

^{2}or Ï€(2r)

^{2}or Ï€d

^{2}

Surface area of a sphere, when diameter of the sphere is

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