Future Study Point

# Class 9 Maths NCERT Solutions For Exercise 13.1 of Chapter 13 -Surface Areas and Volumes

## NCERT Solutions for Class 9 Maths  Chapter 13 Surface Areas and Volumes (Term 2 )CBSE Board exam

Class 9 Maths Exercise-13.1

Class 9 Maths Exercise-13.2

Class 9 Maths Exercise-13.3

Class 9 Maths Exercise -13.5

Class 9 Maths Exercise -13.7

Class 9 Maths Exercise -13.8

Class 9 Maths Exercise-13.9

NCERT Solutions Class 9 Maths-All Chapters

## Class 9 Maths NCERT Solutions For Exercise 13.1 of Chapter 13 -Surface Areas and Volumes

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1.A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open 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 Rs. 20.

Ans. (i)The area of the sheet required for making the box = The surface area of the box which is open at the top

The length of the box is,l = 1.5 m, breadh,b =1.25 m and height,h = 65 cm = 0.65 m

The surface area of the box which is open at the top,S =2(lb + bh + hl) – lb =lb + 2h(l+b)

S = lb + 2h(l+b)

S =1.5×1.25 + 2×0.65(1.5+ 1.25)

S = 1.875 + 1.30(2.75)

S = 1.875 + 3.575

S =5.45

The area of the sheet required for making the box is 5.45 cm²

The cost of sheet measuring 1m2 costs Rs. 20

Therefore the cost of sheet is

20 × 5.45 = Rs 109

Q2. 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 ceiling at the rate of Rs 7.50 per m².

Ans. Surface area four walls and ceiling = 2h(l+b) + lb

l= 5 m, b = 4 m,h = 3m

Surface area of four walls = 2×3(5 +4) +5×4= 6× 9+20 = 54+20= 74m²

The rate of white washing the walls of the room is Rs 7.50 per m²

The cost of white washing the walls of the room is = 74 × 7.50 = Rs 555

Q3.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 m2is Rs.15000, find the height of the hall.

Ans. Let the height of the hall is h

The perimeter of rectangular hall is = 2 (Length + Breadh) =250 m

2(l+b) = 250

Total cost of painting the four walls = Rs 15000

The rate of painting the walls = Rs 10 /m²

Surface area of four walls = 15000/10 = 1500 m²

2h(l+b) = 1500

h 2(l+b) =1500

h× 250 =1500

h = 1500/250 =6

Therefore height of the hall is 6 m

Q4.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. Let the number of bricks to be painted are x

It is given that the paint in a certain container is sufficient to paint an area equal to 9.375 m²

The length ,breadth and height of of a brick is 22.5 cm×10 cm×7.5 cm

Total surface area of a brick is =2(lb +bh +hl)

TSA of a brick = 2(22.5 ×10 +10 ×7.5 + 7.5×22.5)

= 2(225 +75 + 168.75)

=2×468.75 =937.5 cm² = 937.5/10000 =0.09375 m²

The number of bricks to be painted = 9.375 /0.09375 = 100

Q5.A cubical box has each edge 10 cm and another cuboidal box is 12.5cm 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.

(i) The cubical box has the edge of 10 cm and the cuboidal box has length 12.5 cm,breadth 10 cm and height 8 cm

The lateral surface area of cube = 4×side²

LSA of the cubical box= 4×10² = 400 cm²

The lateral surface area of cuboidal box = 2h(l+b)

= 2×8(12.5 +10) = 16×22.5 = 360 cm²

LSA of the cuboidal box= 360 cm²

The lateral surface area of the cubical box is greater than the cuboidal box by (400-360=40 cm²) .

(ii) Total surface area of the cubical box = 6×side² = 6×10² =600 cm²

Total surface area of cuboidal box

= 2(lb + bh + hl)

= 2(12.5 ×10+10×8+8×12.5)

=2(125 +80 + 100) =2×305 =610 cm²

The total surface area of the cubical box is smaller than the cuboidal surface area by (610 -600 )=10 cm²

Q6.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. The length of the greenhouse,l =30 cm, breadth ,b =25 cm and height, h =25 cm

(i) Area of the glass = Total surface area of the greenhouse

TSA of the green house = 2(lb +bh +hl)

= 2(30×25 +25×25 +25×30)

=2(750 + 625 +750)

=2×2125 = 4250 cm²

Hece the area of the glass = 4250 cm²

(ii) The legth of the tape required for 12 edges

The legth of the tape = 4l + 4b + 4h = 4×30 + 4×25 +4×25 =120 +100+100 =320 cm

Q7.Shanti Sweets Stalll 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. The dimension of the bigger box l ×b×h = 25 cm ×20 cm × 5 cm

TSA of the bigger box = 2(lb +bh +hl)

= 2(25×20 +20×5 +5×25)

=2(500 +100 +125)

=2×725 = 1450 cm²

TSA of 250 bigger boxes = 250 ×1450 =362500 cm²

The dimension of the smaller box l ×b×h = 15 cm × 12 cm × 5 cm

TSA of the bigger box = 2(lb +bh +hl)

= 2(15×12 +12×5 +5×15)

=2(180 +60 +75)

=2×315 = 630 cm²

TSA of 250 smaller boxes = 250 ×630 =157500 cm²

TSA of both types of 250 boxes = 362500 +157500 =520000 cm²

Overlaped area of card board = 5 % of 520000 = 26000 cm²

Total area of the cardboard required = 520000 + 26000 =546000 cm²

The cost of 1000 cm² cardboard =Rs 4

The cost of 1 cm² = Rs (4/100)

The cost of the cardboard supplying 250 boxes of each kind = (4/100)×546000 = Rs 2184

Q8.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. Tarpaulin required to cover four sides and top of the car = Lateral surface area of the cover + Top area of the cover

TSA of the cover = 2h(l+b) + lb

l = 4 m, b = 3m, h = 2.5m

TSA of the cover = 2×2.5(4+3) + 4×3

= 5× 7 + 12

=35 + 12

= 47

Hence tarpaulin required for the shelter of the car is 47 m²

You can compensate us by donating any amount of money for our survival

## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions of class 9 science

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

### NCERT Solutions of class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Scroll to Top
Optimized by Seraphinite Accelerator
Turns on site high speed to be attractive for people and search engines.