How to solve NCERT specific maths questions of Mensuration - Future Study Point

How to solve NCERT specific maths questions of Mensuration

How to solve NCERT-specific maths questions of Mensuration

Mensuration is the branch of mathematics in which we study physical quantities like area, perimeter, surface area, and volume of all geometrical figures. The questions of mensuration compulsorily are part of the maths textbook of every class and the questions of mensuration are also asked in every competitive exam  so this post of Future Study Point is useful for everybody. Mensuration is classified into two parts (i) Two-dimensional geometrical figure (ii) Three-dimensional geometrical figure, here we will discuss few specific questions of mensuration for which students generally needed help from his tutor or teacher. These are the specific questions of mensuration which are asked in the CBSE board exam and in the competitive exam. We hope you will definitely get help from the way of solving the questions. Questions are explained through a step by step method by an expert of maths so that everybody could understand the way of solution.

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In this post, we have given 7 examples based on the Mensuration with solutions which are generally asked in various entrance exams. The solutions of these examples will let you know about the way of solution to the mensuration question. For the practice we also have created 8 questions which you can easily solve, if you face any problem then don’t hesitate to write in the comment box, you are free to communicate with us.

NCERT Solutions Class 10 Science from chapter 1 to 16

How to solve NCERT specific maths questions of Mensuration

Example1.The length of a squire field is 25 m, how much area will be of the path of the width 3.5 m around it.

Ans. In this question, a squire field of the length 25 m is given

how to solve mensuration question example 1


A path of the width 3.5 m is given around it given in the second figure

There are two squires one is ABCD which is circumscribed by another squire EFGH

Length of the square EFGH will be = 25 m + 2 × 3.5 = 25 +7 =32 m

Area of path = Area of  squire EFGH – Area of squire ABCD

Area of path = 32 × 32 – 25 × 25

Area of path = 1024 – 625

Area of path = 399 sq.m

Therefore the area of bath = 399 sq.m

In the same way, you can solve all the following questions based on rectangles and circles.

Q1-The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length of each of the straight portions is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.
Q2-Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. And the total cost of developing a garden around the house at the rate of Rs. 55 per
Q3.The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 –  (3.5 + 3.5) metres].

how to solve mensuration questions question 3

Q4. The diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is the same.

Example.2- A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2?

Ans- In this question, we are given the size of tiles which is in the shape of a parallelogram having the base of 24 cm and corresponding height 10 cm and the area of floor the tiles to be installed.

Area of parallelogram = Base × Altitude

Area of tile = 24 × 10 = 240

Let the number of tiles required to cover the floor area = n

n= (Area of the floor)/Area of tile

Because floor area is given in sq.m, therefore, converting it into

1o80 m²

= 1080×100 ×100 = 10800000 cm²

n = (10800000)/240 = 45000

Therefore 45000 tiles are required to cover the floor

Solve the following similar questions

Q5.The floor of a building consists of 3000 tiles which are rhombus shaped end each of its diagonals is 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is Rs. 4.

Q6.What will be the labor charge for tiling a hall 72 m long and 18 m wide at the rate of Rs 12 per sq.m?

Example 3. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. if one of its diagonals is 8 cm long, find the length of the other diagonal.

Ans. Rhombus is a parallelogram have all sides are equal

Area of rhombus = (1/2) product of both diagonals = Side ×Corresponding altitude

Let one of the diagonal is d

(1/2) 8d = 6×4

d = (24/8)×2 = 6

Therefore another diagonal is of the length 6 m

Solve these questions based on rhombus

Q7.The perimeter of a rhombus is 420 meter and one of its diagonal has a length of 40 meters. Find the area of the rhombus.

Q8.The top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Example4-Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth, and height of 15 m, 10 m, and 7 m respectively. From each can of paint 100, m2 of the area is painted.

Ans. The area to painted is- four walls and ceiling, therefore determining the area of four wall and ceiling

Four walls area = 2h(l + b) and area of ceiling = lb

Area of both walls & ceiling = 2h(l + b) + lb

l=15 m, b = 10 m, h = 7 m

= 2 ×7(15 + 10) + 15 × 10

=14 × 25 + 150

= 350 + 150

= 500 sq.m

number of canes reqired to  paint the area = area to be painted/capacity of cane

Number of canes required to paint the given area = 500/100 = 5

Therefore 5 canes are required to  paint the given area

Example5-The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet.

Ans. The lateral surface area of the hollow cylinder is given to us = 4224

The hollow cylinder is cut along its height and formed a rectangular sheet of the width(h) 33 cm and its length would become perimeter of circular ends as shown in the figure.

example 5 how to solve mensuration questions


The lateral surface area of cylinder = 2πrh = lh

lh = 4224

l × 33 = 4224

l = 4224/33 = 128

Therefore the length of rectangular sheet = 128 cm

The perimeter of the rectangular sheet = 2(l + b) =2(l + h)

⇒2(l + h) = 2(128 + 33) = 2×161 = 322

Therefore the perimeter of the rectangular sheet is 322 cm

Example6-A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

Ans. The road roller covers the area of road equal to its curved surface area in 1 revolution = 2πrh

r = 84/2 = 42 cm = 0.42 m

Curved Surface Area of road roller = 2πrh = 2×(22/7)×0.42 ×1 = 2.64


The roller takes 750 revolutions to level the road

∴ The area of road = Number of revolutions taken by the road roller×CSA of road roller750 × 2.64 = 1980

Therefore the area of road = 1980 sq.m

Example7-Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the tank.

example 7 how to solve mensuration questions


Ans. The speed of water pouring into the reservoir = 60 liters/minute.

Converting 60 liters/minute into cubic.meter/minute

⇒ 1 cubic.meter = 1000 l

⇒60 l = 60/1000 = 0.06 cub.meter/minute

∴ The speed of the water in filling the tank = 0.06 cub.meter/minute

Volume of reservoir = 108 cub.meter

Time to be taken to fill the reservoir

=Volume of reservoir/speed of water pouring into it

=108/0.06 = 1800

Therefore the reservoir takes 1800 minutes (30 hours) to fill the tank.

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

NCERT Solutions for 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

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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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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
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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

NCERT Solutions for 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

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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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