# NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1

One of the key subjects for students in Class 9 is Mathematics. It is essential for students to have a foundational understanding of Mathematics at the school level in order to improve their ability to reason and think logically. Mathematical principles may be used by students to address difficulties in the real world. Every industry uses Mathematics in some capacity. A mathematical background can help students advance their careers in a variety of fields. Through the application of its concepts, students can pursue professions in Astronomy, Astrophysics, Statistics, Weather Forecasting, and other related disciplines. Mathematics is one of the core disciplines of a curriculum, it is studied in schools starting in the first grade. The foundational concepts of Mathematics are crucial for a student’s overall growth. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 may be used by students to get a thorough understanding of the mathematical principles discussed in the Class 9 Mathematics curriculum. Students in Class 9 may make use of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 to help them with their preparation for the senior secondary examination in Mathematics. It is essential that students pay attention to the mathematical principles taught in Class 9 since they are used as the foundation for the majority of competitive exams. Since most competitive examinations encapsulate the fundamentals of Class 9 Mathematics, understanding the topics taught in Class 9 Mathematics is essential to appearing in competitive examinations. If students plan to take competitive exams like the NEET, JEE Mains, JEE Advance, CAT, MAT, etc., they should pay particular attention to the mathematical concepts covered in Class 9 in order to do well on such exams. To accomplish the same, students in Class 9 can refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. Students can access the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 from the Extramarks website.

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

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### Access NCERT Solutions for Class 9 Maths Chapter 13- Surface Areas and Volumes

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### NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume

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### NCERT Solutions For Class 9  Maths Chapter 13 Exercise 13.1

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### NCERT Solutions for Class 9

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### CBSE Study Materials for Class 9

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### CBSE Study Materials

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Q.1 A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m2 costs ₹ 20.

Ans

$\begin{array}{l}\text{ Length of the box = 1.5\hspace{0.17em}m}\\ \text{ Breadth of the box = 1.25\hspace{0.17em}m}\\ \text{ Height of the box = 65\hspace{0.17em}cm}\\ \text{ = 0.65\hspace{0.17em}m}\\ \left(\text{i}\right)\text{ Surface area of open box}\\ \text{= lb+2}\left(\text{b+l}\right)\text{h}\\ \text{= 1.5×1.25+2}\left(\text{1.25+1.5}\right)\text{×0.65}\\ \text{= 1.875+3.575}\\ {\text{= 5.45\hspace{0.17em}m}}^{\text{2}}\\ \text{Thus, the area of the sheet required for making the box is}\\ {\text{5.45\hspace{0.17em}m}}^{\text{2}}\text{.}\\ \left(\text{ii}\right){\text{Cost of 1m}}^{\text{2}}\text{sheet = ₹\hspace{0.17em}20}\\ \text{Cost of sheet = ₹\hspace{0.17em}20×5.45 = ₹109}\end{array}$

Q.2 The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m2.

Ans

$\begin{array}{l}\text{The length of a room = 5\hspace{0.17em}m}\\ \text{The breadth of a room = 4\hspace{0.17em}m}\\ \text{The height of a room = 3\hspace{0.17em}m}\\ \text{Surface area of 4 walls and ceiling = lb+2}\left(\text{b+l}\right)\text{h}\\ \text{\hspace{0.17em} = 5×4+2}\left(\text{4+5}\right)\text{×3}\\ \text{ \hspace{0.17em}= 20+54}\\ {\text{ = 74\hspace{0.17em}m}}^{\text{2}}\\ {\text{Rate of whitewashing of 1 m}}^{\text{2}}\text{area = ₹\hspace{0.17em}7.50}\\ \text{Rate of whitewashing the room = ₹\hspace{0.17em}7.50×74}\\ \text{ = ₹ 555}\\ \text{Thus, the cost of whitewashing the room is ₹ 555.}\end{array}$

Q.3 The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000, fin the height of the hall.

Ans

$\begin{array}{l}\text{The perimeter of a rectangular hall = 250\hspace{0.17em}m}\\ {\text{ Rate of painting the walls = ₹10 per m}}^{\text{2}}\\ \text{ The cost of painting the 4 walls = ₹15000}\\ \text{ Let height of hall = h\hspace{0.17em}m}\\ \text{ Area of 4 walls of the room = 2}\left(\text{l+b}\right)\text{h}\\ \text{ = 250\hspace{0.17em}h}\\ \text{ Cost of painting the four walls = 10×250\hspace{0.17em}h}\\ \text{ ₹ 15000 = 10×250\hspace{0.17em}h}\\ \text{ h = }\frac{\text{15000}}{\text{2500}}\text{ = 6\hspace{0.17em}m}\\ \text{Thus, the height of hall is 6 m.}\end{array}$

Q.4 The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Ans

$\begin{array}{l}{\text{The area to paint by a certain container = 9.375\hspace{0.17em}m}}^{\text{2}}\\ \text{Dimensions of given brick\hspace{0.17em}= 22.5cm×10cm×7.5cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Surface area of a brick = 2}\left(\begin{array}{l}\text{22.5cm×10cm+10cm×7.5cm}\\ \text{+7.5cm×22.5cm}\end{array}\right)\\ \text{ = 2}\left(\text{225+75+168.25}\right)\\ \text{ = 2}\left(\text{468.75}\right)\\ {\text{ = 937.5\hspace{0.17em}cm}}^{\text{2}}\\ \text{ = }\frac{\text{937.5}}{\text{10000}}{\text{\hspace{0.17em}m}}^{\text{2}}\\ \text{Number of bricks painted in given container = }\frac{\text{9.375}}{\left(\frac{\text{937.5}}{\text{10000}}\right)}\\ \text{ = 100}\\ \text{Thus, 100 bricks can be painted out of this container.}\end{array}$

Q.5 A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

(i) Which box has the greater lateral surface area and by how much?

(ii) Which box has the smaller total surface area and by how much?

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Edge of a cubical box}=10\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Lateral\hspace{0.17em}}\mathrm{surface}\text{area of cubical box}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}=4{\left(10\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=400\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length of cuboidal box}=\text{\hspace{0.17em}}12.5\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Breadth of cuboidal box}=\text{\hspace{0.17em}}10\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Height of cuboidal box}=\text{\hspace{0.17em}}8\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Lateral\hspace{0.17em}}\mathrm{surface}\text{area of cuboidal box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}2\left(\mathrm{l}+\mathrm{b}\right)\mathrm{h}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(12.5+10\right)×8\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=360\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{cubical box has the greater lateral surface area.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Greater area of cubical box}=400-360\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=40\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{(ii) Total surface area of cubical box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{\left(\mathrm{side}\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{\left(10\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=600\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Total\hspace{0.17em}}\mathrm{surface}\text{area of cuboidal box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}2\left(\mathrm{lb}+\mathrm{bh}+\mathrm{hl}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(12.5×10+10×8+8×12.5\right)\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(125+80+100\right)\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(305\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=610\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Total\hspace{0.17em}surface area of cuboidal box is greater than cubical box.}\\ \text{Greater surface area of cubical box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=610-600\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\end{array}$

Q.6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

Ans

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

Q.7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

Ans

Total surface area of bigger box
= 2(lb + bh + hl)
= 2(25×20+ 20×5+ 5×25)
= 2 (725)
= 1450 cm2
Required extra surface area = 5% of 1450 cm2
=72.5 cm2
Total surface area of smaller box
= 2(lb + bh + hl)
= 2(15×12+12×5+ 5×15)
= 2 (315)
= 630 cm2
Required extra surface area = 5% of 630 cm2
=31.5 cm2
Total required cardboard for both types boxes
= 1450 + 72.5 + 630 + 31.5
= 2184 cm2
Total required cardboard for both types 250 each boxes

= 250 x 2184 cm2
= 546000 cm2
Cost of 1000 cm2 cardboard = Rs. 4
Cost of required cardboard = Rs. (4/1000) x 546000
= Rs. 2184

Q.8 Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

Ans

The length of shelter = 4 m
The breadth of shelter = 3 m
The height of shelter = 2.5 m
Total surface area of shelter of car
= lb + 2(b + l) × h
= 4 × 3 + 2(3+ 4)× 2.5 m2
= 12 + 35
= 47 m2
Thus, the required tarpaulin for shelter of car is 47 m2.