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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

The NCERT has precisely designed the solutions in such a way that it teaches students about the various concepts of Geometry, along with the application of relevant formulas. NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.9 help in solving the application level complex problems in an easy way.

NCERT solutions are the result of the combined effort of our subject-matter experts and teaching faculty, who have strived to make the concept understandable. NCERT solutions will ensure that the answers to the questions are according to the latest syllabus. It is taken care of to provide solutions in simple language with understandable examples. NCERT solutions provide an easy way for students to prepare for the board exams, as they will be more familiar with the questions of the exercise.

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Access Other Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

Exercise 13.1 solution (9 questions)

Exercise 13.2 solution (8 questions)

Exercise 13.3 solution (9 questions)

Exercise 13.4 solution (5 questions)

Exercise 13.5 solution (5 questions)

Exercise 13.6 solution (8 questions)

Exercise 13.7 solution (9 questions)

Exercise 13.8 solution (10 questions)

Access Answers to NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.9

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 class 9 chapter 13-20

Solution:

The external dimensions of the bookshelf,

Length, l = 85cm

Breadth, b = 25 cm

Height, h = 110 cm

The external surface area of the shelf while leaving out the front face of the shelf

= lh+2(lb+bh)

= [85×110+2(85×25+25×110)] = (9350+9750) = 19100

The external surface area of the shelf is 19100 cm2

The area of front face = [85×110-75×100+2(75×5)] = 1850+750

So, the area is 2600 cm2

The area to be polished = (19100+2600) cm2 = 21700 cm2

The cost of polishing 1 cm2 area = Rs 0.20

The cost of polishing 21700 cm2 area Rs. (21700×0.20) = Rs 4340

The dimensions of the row of the bookshelf

Length (l) = 75 cm

Breadth (b) = 20 cm and

Height (h) = 30 cm

The area to be painted in one row = 2(l+h)b+lh = [2(75+30)× 20+75×30] = (4200+2250) = 6450

So, the area is 6450 cm2.

The area to be painted in 3 rows = (3×6450)cm2 = 19350 cm2

The cost of painting 1 cm2 area = Rs. 0.10

The 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

Answer: The cost for polishing and painting the surface of the bookshelf is Rs. 6275.

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 the 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 class 9 chapter 13-21

Solution:

The diameter of the wooden sphere = 21 cm

The radius of the wooden sphere, r = diameter/ 2 = (21/2) cm = 10.5 cm

Formula: The surface area of the wooden sphere = 4πr2

= 4×(22/7)×(10.5)2 = 1386

So, the surface area is 1386 cm3

The radius of the circular end of cylindrical support = 1.5 cm

The height of cylindrical support = 7 cm

The curved surface area = 2πrh

= 2×(22/7)×1.5×7 = 66

So, CSA is 66 cm2

Now,

The area of the circular end of cylindrical support = πr2

= (22/7)×1.52

= 7.07

The area of the circular end is 7.07 cm2

Again,

The area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44

The area to be painted is 11031.44 cm2

The cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86

The area to be painted black = (8×66) cm2 = 528 cm2

The 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

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”.

The radius of the sphere, r1 = d/2

The new radius of the sphere, say r2 = (d/2)×(1-25/100) = 3d/8

The curved surface area of sphere, (CSA)1 = 4πr12 = 4π×(d/2)2 = πd2 …(1)

The curved surface area of the sphere when the radius is decreased (CSA)2 = 4πr22 = 4π×(3d/8)2 = (9/16)πd2 …(2)

From equations (1) and (2), we have

Decrease in surface area of sphere = (CSA)1 – (CSA)2

= πd2 – (9/16)πd2

= (7/16)πd2

Ncert solutions class 9 chapter 13-22

= (7d2/16d2)×100 = 700/16 = 43.75% .

Therefore, the percentage decrease in the surface area of the sphere is 43.75% .


A more in-depth explanation with an accurate solution can be accessed from the NCERT chapter-wise solutions that we have provided. The 3 questions asked in this exercise are long answer questions and are optional according to the syllabus. The NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.9 is aimed at helping students score good scores.

Solving exercise problems chapter-wise will help them

  • Get sufficient practice to solve the questions confidently.
  • Get an overall understanding of the concepts.
  • It helps them to be familiar with the different types of questions that appear in the exam.
  • It helps to increase their accuracy and speed in solving problems.

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