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

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

class 9 math ncert chapter 13.1

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

class 9 math ncert chapter 13.1

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

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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

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

NCERT Solutions of class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions of class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

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NCERT Solutions of class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

Solutions of class 10 last years Science question papers

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CBSE class 10 -Latest sample paper of science

NCERT solutions of class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 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 FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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