NCERT Solutions for Exercise 13.9 Class 9 Maths Chapter 13

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

Edited By Sumit Saini | Updated on Jul 06, 2022 02:31 PM IST

NCERT Solutions for Class 9 Maths exercise 13.9 is an optional last exercise which covers all of the topics from the whole chapter, including Surface Areas And Volumes of cylinder, sphere, etc…. In exercise 13.9 Class 9 Maths, The cuboid is a three-dimensional figure defined by six rectangular planes with different magnitudes of length, width, and height. It has eight vertices and twelve edges, and opposite faces are always equal. Also a sphere is a three-dimensional object that is circular in shape. The cuboid has two different types of surface area. They are lateral surface area and total surface area. Similarly, The surface area of a sphere refers to the overall area covered by its surface in three-dimensional space.

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Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.9
More About NCERT Solutions for Class 9 Maths Exercise 13.9
Benefits of NCERT Solutions for Class 9 Maths Exercise 13.9
NCERT Solutions of Class 10 Subject Wise

NCERT solutions for Class 9 Maths chapter 13 exercise 13.9 include three questions that are not crucial for the examination but will help to improve geometrical concepts and creative thinking skills, which will be a useful addition to mathematical knowledge. The following exercises are also present along with NCERT book Class 9 Maths chapter 13 exercise 13.9.

Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.9

Q1 A wooden bookshelf has external dimensions as follows: Height $\small =110\hspace{1mm}cm$ , Depth $\small =25\hspace{1mm}cm$ , Breadth $\small =85\hspace{1mm}cm$ (see Fig. $\small 13.31$ ). The thickness of the plank is $\small 5\hspace{1mm}cm$ everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $\small 20\hspace{1mm}$ paise per $\small cm^2$ and the rate of painting is $\small 10\hspace{1mm}$ paise per $\small cm^2$ , find the total expenses required for polishing and painting the surface of the bookshelf.

1640848333224

Answer:

External dimension od the bookshelf = $85\ cm\times25\ cm\times110\ cm$

(Note: There is no front face)

The external surface area of the shelf = $lh + 2 (lb + bh)$
$\\ = [85 \times 110 + 2 (85 \times 25 + 25 \times 110)] \\ = (9350 + 9750) \\ = 19100\ cm^2$

We know, each stripe on the front surface is also to be polished. which is 5 cm stretch.

Area of front face = $[85 \times 110 - 75 \times 100 + 2 (75 \times5)]$

$\\ = 1850 + 750 \\ = 2600\ cm^2$

Area to be polished = $(19100 + 2600) = 21700\ cm^2$

Cost of polishing $1\ cm^2$ area = $Rs\ 0.20$

Cost of polishing $21700\ cm^2$ area = $Rs.\ (21700 \times 0.20) = Rs.\ 4340$

Now,

Dimension of inner part = $75\ cm\times15\ cm\times100\ cm$

Area to be painted in 3 rows = $3\times[2 (l + h) b + lh]$

$\\ =3\times [2 (75 + 30) \times 20 + 75 \times 30] \\ = 3\times[(4200 + 2250)] \\ = 3\times6450 \\ = 19350\ cm^2$

Cost of painting $1\ cm^2$ area = $Rs\ 0.10$

Cost of painting $19350\ cm^2$ area = $Rs.\ (19350 \times 0.10)= Rs.\ 1935$

Total expense required for polishing and painting = $Rs.\ (4340 + 1935)$

$= Rs.\ 6275$

Q2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig $\small 13.32$ . Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius $\small 1.5\hspace{1mm}cm$ and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per $\small cm^2$ and black paint costs 5 paise per $\small cm^2$ .

1640848312910

Answer:

Given,

The radius of the wooden spheres = $r_1 = \frac{21}{2}\ cm$

$\therefore$ The surface area of a single sphere = $4\pi r_1^2$

$\\ = 4\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2} \\ = 22\times3\times21$

$= 1386\ cm^2$

Again, the Radius of the cylinder support = $\small r_2 = 1.5\hspace{1mm}cm$

Height of the support = $h = 7\ cm$

$\therefore$ The base area of the cylinder = $\pi r_2^2 = \frac{22}{7}\times1.5\times1.5 = 7.07\ cm^2$

Now, Cost of painting $1\ cm^2$ silver = $25\ paise = Rs.\ 0.25$

$\therefore$ Cost of painting 1 wooden sphere = Cost of painting $(1386-7.07)\ cm^2$ silver

= $Rs. (0.25\times1378.93) = Rs.\ 344.7325$

Now, Curved surface area of the cylindrical support = $2\pi r_2 h$

$\\ = 2\times\frac{22}{7}\times1.5\times7 \\ = 22\times3 \\ = 66\ cm^2$

Now, Cost of painting $1\ cm^2$ black = $5\ paise = Rs.\ 0.05$

$\therefore$ Cost of painting 1 such stand = Cost of painting $66\ cm^2$ silver = $Rs. (0.05\times66) = Rs.\ 3.3$

$\therefore$ The total cost of painting 1 sphere and its support = $Rs.\ (344.7325+3.3) = Rs.\ 348.0325$

Therefore, total cost of painting 8 such spheres and their supports = $Rs.\ (8\times348.0325) = Rs.\ 2784.26$

Q3 The diameter of a sphere is decreased by $\small 25\%$ . By what per cent does its curved surface area decrease?

Answer:

Let the radius of the sphere be $r$

Diameter of the sphere = $2r$

According to question,

Diameter is decreased by $\small 25\%$

So, the new diameter = $\frac{3}{4}\times2r = \frac{3r}{2}$

So, the new radius = $r' = \frac{3r}{4}$

$\therefore$ New surface area = $4\pi r'^2 = 4\pi (\frac{3r}{4})^2$

$\therefore$ Decrease in surface area = $4\pi r^2 - 4\pi (\frac{3r}{4})^2$

$= 4\pi r^2[1 -\frac{9}{16} ]$

$= 4\pi r^2[\frac{7}{16} ]$

$\therefore$ Percentage decrease in the surface area = $\frac{Difference\ in\ areas}{Original\ surface\ area}$

$= \frac{4\pi r^2[\frac{7}{16}]}{4\pi r^2}\times100 \%$

$= \frac{7}{16}\times100 \%$

$= 43.75 \%$

Benefits of NCERT Solutions for Class 9 Maths Exercise 13.9

• NCERT solutions for Class 9 Maths exercise 13.9 helps in solving the application level complex problem in an easy way and get sufficient practice to solve the questions confidently.

• By solving the NCERT solution for Class 9 Maths chapter 13 exercise 13.9 exercises, we can get an overall understanding of the concepts and helps us to be familiar with the different types of questions in the exam

• Exercise 13.9 Class 9 Maths, helps to increase your accuracy and speed in solving problems.

Also, See:

NCERT Solutions of Class 10 Subject Wise

Frequently Asked Questions (FAQs)

1. A cuboid's lateral surface area is ________

A cuboid's lateral surface area is 2h(l + b)

2. The cuboid's total surface area is _______

TSA = 2(lb + bh + hl) is the total surface area of a cuboid .

3. The sphere's surface area is _________

The sphere's surface area is 4πr^2 .

4. The surface area of the cube with the edge m is ______

The surface area of the cube with the edge m is 6m^2

5. Define cube.

A cube is a three-dimensional shape with the same length, breadth, and height as it has six faces, eight vertices, and twelve edges.

6. Define the radius of the sphere using the NCERT solutions for Class 9 Maths chapter 13 exercise 13.9 .

NCERT solutions for Class 9 Maths chapter 13 exercise 13.9 are as follows: The distance between a sphere's surface and its centre is its radius.

7. Write the relationship between radius and diameter.

The radius of a circle is twice its diameter . 2×radius = diameter

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