• NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • Class 6-10
      • Class 6th
      • Class 7th
      • Class 8th
      • Class 9th
      • Class 10th
    • View All Options
      • Online Courses
      • Offline Courses
      • Distance Learning
      • Hindi Medium Courses
      • International Olympiad
    • NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE (Main+Advanced)
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE Main
      • Class 11th
      • Class 12th
      • Class 12th Plus
  • NEW
    • JEE 2025
    • NEET
      • 2024
      • 2023
      • 2022
    • Class 6-10
    • JEE Main
      • Previous Year Papers
      • Sample Papers
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • JEE Advanced
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NEET
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NCERT Solutions
      • Class 6
      • Class 7
      • Class 8
      • Class 9
      • Class 10
      • Class 11
      • Class 12
    • CBSE
      • Notes
      • Sample Papers
      • Question Papers
    • Olympiad
      • NSO
      • IMO
      • NMTC
    • TALLENTEX
    • AOSAT
    • ALLEN e-Store
    • ALLEN for Schools
    • About ALLEN
    • Blogs
    • News
    • Careers
    • Request a call back
    • Book home demo
NCERT SolutionsCBSE NotesCBSE Exam
Home
Maths
Mensuration

Mensuration

Different shapes

1.0Standard Units Of Volume

 Value of π=722​ or 3.141dm=10 cm 1dm3 or litre =1dm×1dm×1dm=(10×10×10)cm3=1000 cm3 1 cm=10 mm 1 m=100 cm 1 m=10dm 1 m3=(100×100×100)cm3=1000000 cm3 1 m3=1000×1000 cm3=1000 litre 1 kilolitre =1 m3=1000 litre (mm) millimetre, (cm) centimetre, (dm) decimetre, (m) metre

mensuration of different shapes

2.0Plane Figures

The geometrical figures which have only two dimensions are called as the plane figures i.e. Triangle, Square, Parallelogram, Trapezium etc.

Examples of Plane figure

Take a glass partly filled with water and pour some oil in it.

Image of glass partly filled with water

Observe : Do the surfaces of oil and water have any height? No, that is why that surface is called two-dimensional surface because it does not have height.

3.0Solid Figures

A figure which has three dimensions (as length, breadth and height) is not a plane figure and we cannot draw such figures on black board exactly. These three-dimensional figures are called solids. Three dimensions or 3D are length, breadth and height. E.g., Cube, Cuboid, Cylinder, Cone, Sphere, Prism, Pyramid etc.

Cube-a solid figue

Cube

Cylinder-a solid figure

Cylinder

Prism

Prism

Cone

Cone

4.0Triangle

(i) Area of triangle =21​× Base × Altitude or 21​(b×h)

Area of a triangle

(ii) Area of an equilateral triangle =43​​( side )2 or 43​​a2

Area of equilateral triangle

(iii) Area of an isosceles triangle (base =b, equal side =a ) =4 base ​×4( equal side )2−( base )2​ or 4b​×4a2−b2​

Area of isosceles triangle

  • Q. Find the area of each triangle below.

Sample problem triangle

  • (a)
    (b) Solution : Area of triangle =21​×b×h or 21​× base × height (a) Area =21​×5×4.2=10.5 cm2 (b) Area =21​×6×5.5=16.5 cm2
  • Q. Calculate the area of an equilateral triangle whose perimeter is 24 m . Solution :
    Perimeter of an equilateral triangle =3a=24 m {" a " is the side of equilateral triangle } ⇒3a=24 ⇒a=8 Thus, the length of side is 8 m . Area of equilateral triangle =43​​×( side )2 =43​​×82 =163​ m2
  • Q. Calculate the area of an isosceles triangle whose sides are 13 cm,13 cm and 24 cm . Explanation : Equal side (a) = 13 cm base (b) =24 cm Area of isosceles triangle =4b​4a2−b2​ =424​4(13)2−(24)2​=6676−576​=6100​=6×10=60 cm2

5.0Quadrilateral

Area of quadrilateral ABCD=21​×AC×(h1​+h2​) =21​× (length of a diagonal) × (sum of the lengths of perpendiculars from the remaining two vertices on it)

Here, h1​ and h2​ are called the offsets of the quadrilateral.

  • Q. The diagonal of a quadrilateral is 20 m in length and the perpendiculars to it from the opposite vertices are 8.5 m and 11 m . Find the area of the quadrilateral.
    Solution : In quadrilateral ABCD, we have AC=20 m Let BL⊥AC and DM⊥AC such that BL=8.5 m and DM=11 m ∴ Area of quadrilateral ABCD=21​×AC×(BL+DM) =21​×20×(8.5+11)=10×19.5=195 m2
  • Perimeter is the length of boundary of a closed figure. We can find perimeter by doing sum of all sides.

6.0Rectangle

Length =ℓ, Breadth =b (i) Perimeter of rectangle =2(ℓ+b) (ii) Area of rectangle =ℓ×b (iii) Length = Breadth  Area ​ or Breadth = Length  Area ​ (iv) Diagonal =ℓ2+b2​

Area of path (shaded region) = Area of outer rectangle - Area of inner rectangle

Area of path (shaded region) = Ar. ( ABCD )− Ar. (PQRS)

  • We can find length of diagonal of rectangle by using Pythagoras theorem.
  • Q. The perimeter of a rectangular sheet is 120 cm . If the length is 35cm, find its breadth. Also find the area. Explanation : Perimeter of a rectangular sheet =120 cm (given) Length =35 cm. Perimeter of rectangle =2(ℓ+b) 120=2(35+b) ⇒60=35+b ⇒b=25 cm ∴ Breadth of rectangle =25 cm Area of rectangle =ℓ×b=35×25=875 cm2

7.0Square

Length of each side =a (i) Perimeter of square =4 a or 4× side (ii) Area of square =a2 or (side × side) or 21​( Diagonal )2 (iii) Side of the square = Area ​ (iv) Diagonal of the square =2​a

  • Same as rectangle, we can find the length of diagonal of a square by using Pythagoras theorem.
  • Q. Find the area of the figure shaded in the diagram.
    Explanation : We can see from the diagram that ABCD is a rectangle and PQRS is a square. Area of rectangle ABCD= length × breadth =8 cm×6 cm =48 cm2 Area of square PQRS= side × side =2 cm×2 cm =4 cm2 Area of shaded figure = Area of rectangle ABCD− Area of square PQRS =(48−4)cm2 =44 cm2
  • Q. Find the perimeter and area of each shape below. (a)
    (b)
    Solution : (a) The perimeter is found by adding the length of all the sides. Perimeter =6+8+1+4+4+4+1+8=36 cm To find the area, consider the shape being split into a rectangle and a square
    Area = Area of rectangle + Area of square =(6×8)+42=48+16=64 cm2. (b) Adding the length of the sides gives P=10+7+8+2+2+5=34 cm The area can be found by considering the shape to be a rectangle with a square removed from it. Area of shape = Area of rectangle - Area of square =(7×10)−22=70−4=66 cm2
  • Q. How many square tiles of side 10 cm will be required to tile a floor measuring 800 cm by 900 cm ? Solution: Number of square tiles = Area of a square tile  Area of rectangular floor ​=10×10800×900​=7200 tiles
  • While solving questions, remember that the units must be same.
  • Trapezium is a quadrilateral not a parallelogram as it has only one pair of parallel sides.

8.0Trapezium

A trapezium is a quadrilateral in which one pair of opposite sides are parallel.

Base

Each of the two parallel sides of a trapezium is called base of the trapezium.

Height or altitude

The distance between the two bases (parallel sides) is called the height or altitude of the trapezium.

h is the height of the trapezium ABCD or CL=h Join AC; clearly AC divides the trapezium ABCD into two triangles ABC and ACD. Area of trapezium ABCD= Area of △ABC+ Area of △ACD ′h ' is the altitude for both △ABC & △ACD. Area of △ABC=21​×AB×h and Area of △ACD=21​×DC×h Substituting these values in eq n . (i), we get Area of trapezium ABCD=(21​×AB×h)+(21​×DC×h)=21​(AB+DC)×h Area of trapezium =21​× (Sum of the parallel sides) × (Distance between parallel sides)

  • When we talk about the shortest distance, it is the perpendicular distance from a point to a line or between two lines.
  • Q. The area of a trapezium shaped field is 480 m2, the height is 15 m and one of the parallel sides is 20 m. Find the other side.
    Solution : We have, Area of trapezium ABCD =480 m2 ⇒21​(AB+CD)×AL=480 ⇒21​(20+CD)×15=480 ⇒20+CD=15480×2​ ⇒20+CD=32×2 ⇒20+CD=64 ⇒CD=64−20=44 Hence, the other side of the trapezium is 44 m .
  • Q. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to the given base is 9 cm long. Solution : Given Length of bases of trapezium =15 cm and 9 cm . Length of height =8 cm Area of trapezium =21​×( sum of parallel sides )× height =21​×(15+9)×8=24×4=96 cm2

9.0Parallelogram

(i) Area of parallelogram = Base × Height or AB×DM

(ii) Perimeter =2 (sum of two adjacent sides) or 2(AB+AD)

  • A parallelogram is a quadrilateral in which both pairs of sides are parallel.
  • Q. The parallel sides of a trapezium are 20cm and 10 cm . Its non-parallel sides are both equal, each being 13 cm . Find the area of the trapezium. Explanation :
    Let ABCD be a trapezium such that AB=20 cm,CD=10 cm and AD=BC=13 cm Draw CL∣∣AD and CM⊥AB. Now, CL|AD and CD|AL. ∴ ALCD is a parallelogram. ⇒AL=CD=10 cm and CL=AD=13 cm In △CLB, we have CL=CB=13 cm ∴Δ CLB is an isosceles triangle. LM=MB=21​BL=21​×10 cm=5 cm[∵BL=AB−AL=(20−10)cm=10 cm] Applying Pythagoras theorem in △ CML, we have CL2=CM2+LM2 132=CM2+52 CM2=169−25=144 CM=144​=12 cm ⇒ Area of △CLB=21​×BL×CM=21​×10×12=60 cm2 Area of parallelogram ALCD =AL×CM=(10×12)=120 cm2 Hence, Area of trapezium ABCD = Area of parallelogram ALCD + Area of △ CLB =(120+60)cm2=180 cm2
  • Q. The base of a parallelogram is thrice its height. If the area is 867 cm2, find the base and the height of the parallelogram. Solution : Let the height of the parallelogram be xcm . Then, base =3xcm Area of the parallelogram =(x×3x)cm2=3x2 cm2 But area of the parallelogram is given as 867 cm2 ∴3x2=867 ⇒x2=289 ⇒x=289​=17 Height =17 cm and base =(3×17)cm=51 cm.

10.0Rhombus

If d1​ and d2​ are the lengths of the diagonals of a rhombus, then

(i) Area =21​d1​d2​ or base × height (ii) Perimeter =2 d12​+d22​​ or 4 (side) (iii) Side of the rhombus =21​ d12​+d22​​

  • A rhombus is a parallelogram in which all the sides are equal.
  • Rhombus is the quadrilateral which is not a regular polygon but its all sides are equal.
  • Q. If the area of a rhombus be 24 cm2 and one of its diagonals be 4cm, find the perimeter of the rhombus.
    Explanation : Let ABCD be a rhombus such that its one diagonal AC=4 cm. Suppose the diagonals AC and BD intersect at 0 . Area of rhombus ABCD=24 cm2 ⇒21​×AC×BD=24⇒21​×4×BD=24 ⇒2×BD=24⇒BD=12 cm Thus, we have AC=4 cm and BD=12 cm ∴OA=21​AC=2 cm and OB=21​BD=6 cm Since the diagonals of a rhombus bisect each other at right angle. Therefore, △OAB is right triangle, right angled at 0. Using Pythagoras theorem in △OAB, we have AB2=0 A2+0 B2 AB2=22+62=40 AB=40​=210​ cm Hence, perimeter of rhombus ABCD=(4×210​)cm=810​ cm
  • Q. Find the area of rhombus (given).
    Solution : Using formula, A=21​d1​×d2​ A=21​×5×8=20 cm2.
  • The value of π is 722​ or 3.14.

11.0Circle

(i) Circumference of circle =2πr (ii) Radius of circle =2π Circumference ​ or π Area ​​ (iii) Area of circle =πr2

  • Draw a circle of any radius r. Divide it into 16 parts by paper folding and join them as shown in the figure. Exploring the concept
    The figure looks like a rectangle. Length of this rectangle = length of arc of a semicircle of radius r =21​×2πr=πr and, Breadth of rectangle = radius of circle =r. Drawing conclusion: ∴ Area of circle = area of rectangle =πr×r=πr2.
  • Q. Find the circumference and area of this circle (π=3.14).
    Solution : The circumference is found using C=2πr, which in this case gives C=2π×4=2×3.14×4=25.12 cm The area is found using A=πr2, which gives A=π×42=3.14×16=50.24 cm2

Q. Find the radius of a circle if [π=722​]; (i) its circumference is 32 cm, (ii) its area is 14.3 cm2. Solution : (i) Using C=2πr gives 32=2πr and dividing by 2π gives 2π32​=r r=2×2232×7​=1156​=5.09 cm (ii) Using A=πr2 gives 14.3=πr2 ⇒π14.3​=r2 or r=π14.3​​=2214.3×7​​=4.55​=2.13 cm

12.0Area of a polygon

Polygon

A simple closed plane figure made by three or more line segments is called a polygon.

Regular polygon

A polygon is said to be regular when its all sides and all angles are equal.

Types of polygon

Number of SidesName of Polygon
4Quadrilateral
5Pentagon
6Hexagon
7Heptagon
8Octagon
9Nonagon
10Decagon
  • Q. Find the area of the pentagon ABCDE shown in figure if AD=8 cm,AH=6 cm,AG=4cm, AF=3 cm,BF=2 cm,CH=3 cm and EG=2.5 cm. Solution : We have
    Area of pentagon ABCDE= Area of △AFB+ Area of trapezium FBCH+ Area of △CHD+ Area of △ADE =21​(AF×BF)+21​(BF+CH)×FH+21​(DH×CH)+21​(AD×GE) ⇒21​(AF×BF)+21​(BF+CH)×(AH−AF)+21​{(AD−AH)×CH}+21​(AD×GE) ⇒2(3×2)+(2+3)×(6−3)+(8−6)×3+(8×2.5)​ ⇒26+15+6+20​ ⇒247​=23.5 cm2

13.0Cuboid

A solid bounded by six rectangular plane regions is called a cuboid.

Cuboid

A cuboid has six faces OAQB, CMPN, OAMC, BQPN, PQAM and OCNB, eight vertices A,B,C,O,P,Q, M and N, twelve edges : OA,BQ,NP,CM,PM,CN,OB,AQ,BN,PQ,MA and OC and four diagonals OP, BM, CQ and AN.

Surface area of a cuboid

Consider a cuboid whose length is ℓcm, breadth is bcm and height is h cm as shown in figure. Area of face BNPQ = Area of face AOCM =ℓ×b Area of face BNCO = Area of face AQPM =b×h Area of face CMPN = Area of face AQBO=ℓ×h ∴ Total surface area of the cuboid = sum of the areas of all its six faces =2(ℓ×b)+2( b×h)+2(ℓ×h)=2[ℓ b+bh+ℓh]

Lateral surface area of a cuboid (Area of 4 walls) = Perimeter of the base × Height =2×( Length + breadth )× Height =2(ℓ+b)×h

Volume of cuboid = Area of a rectangular sheet ×h = Length × Breadth × Height  =(ℓ×b)×h=ℓbh

Diagonal of a cuboid =ℓ2+b2+h2​

  • Q. The diagram shows a lorry.
    Find the volume of the load-carrying part of the lorry. Explanation : The load-carrying part of the lorry is represented by a cuboid, so its volume is given by V=2×2.5×4=20 m3.
  • Q. There are two cuboidal boxes whose dimensions are given below. Which box requires the higher amount of material to make? Cuboid A : ℓ=23, b=30,h=40 Cuboid B : ℓ=30, b=12, h=44 Solution : Required amount of material will be measured by the total surface area. Total surface area of cuboidal box A=2(ℓ b+bh+ℓh) =2(23×30+30×40+23×40)=2×2810 cm2=5620 cm2 Total surface area of cuboidal box B=2(ℓb+bh+ℓh) =2(30×12+12×44+30×44)=2×2208 cm2=4416 cm2 So, cuboidal box A will require the higher amount of material to make.

14.0Cube

A cuboid whose length, breadth and height are all equal is called a cube. Since all the faces of a cube are squares of the same size i.e. for a cube we have ℓ=b=h. Thus, if ℓcm is the length of the edge of side of a cube, then Total Surface area of the cube =2(ℓ×ℓ+ℓ×ℓ+ℓ×ℓ) =2×3ℓ2=6ℓ2=6( Edge )2

Lateral surface area of the cube =2(ℓ×ℓ+ℓ×ℓ) =2(ℓ2+ℓ2)=4ℓ2=4( Edge )2

Volume of a cube =ℓ×ℓ×ℓ=ℓ3

  • When the length of the edge of a cube is doubled, its surface area becomes four times and volume becomes eight times.
  • Q. If the volume of a cube is 729 cm3, find its side. Solution Volume of cube =( side )3 729=( side )3⇒3729​= side ⇒393​= side ⇒9= side Hence, side of cube =9 cm
  • Q. Find the side of a cube whose surface area is 2400m2. Explanation : Total surface area of cube =6 (side) 2 ⇒6 (side) 2=2400 ⇒( side )2=400 ⇒ side =400​=20 m Right circular cylinder
    A solid generated by the revolution of a rectangle about one of its side is called a right circular cylinder. Radius of right circular cylinder is r and height is h Area of the lateral surface of the cylinder = Area of the paper strip ABCD =2πr×h Lateral (curved) surface area of cylinder =2πrh=( Base perimeter ×h) Area of each base of cylinder =πr2 Total surface area of cylinder = Area of the curved surface +2× Area of base =2πrh+2πr2=2πr(h+r) Thus, volume of right circular cylinder = Area of the base × height =πr2h
  • Q. A milk tank is in the form of cylinder whose radius is 1.5 m and length 7m. Find the quantity of milk in litres that can be stored in the tank.
    Solution : Radius ( r ) =1.5 m, Height ( h ) =7 m, Volume of cylinder =πr2 h =(722​×1.5×1.5×7)m3=49.5 m3 ∵1 m3=1000 litres ∴49.5 m3=1000×49.5 litres =49500 litres Hence, the required quantity of milk in litres in tank =49500 litres
  • Q. The earth taken out while digging a pit, is evenly spread over a rectangular field of length 90 m , width 60 m . If the volume of the earth dug is 3078 m3, find the height of the field raised. Explanation : 3078 m3=90 m×60 m×h h=90×603078​=0.57 m ∴ Height of field raised =0.57 m=0.57×100=57 cm(∵1 m=100 cm)

15.0Mind Map

Table of Contents


  • 1.0Standard Units Of Volume
  • 2.0Plane Figures
  • 3.0Solid Figures
  • 4.0Triangle
  • 5.0Quadrilateral
  • 6.0Rectangle
  • 7.0Square
  • 8.0Trapezium
  • 8.1Base
  • 8.2Height or altitude
  • 9.0Parallelogram
  • 10.0Rhombus
  • 11.0Circle
  • 12.0Area of a polygon
  • 13.0Cuboid
  • 14.0Cube
  • 15.0Mind Map

Related Articles:-

Understanding Elementary Shapes

When we measure a line segment, we measure its length or distance from one end point to the other.

Decimals

Decimals are another way of writing part of a whole number. In a place value table on moving a digit...

Integers

As we know that when a smaller whole number is subtracted from larger whole number we get a whole number...

Whole Numbers

The number which comes immediately after a particular number is called its successor. The successor...

Playing with Numbers

To understand the concept of prime and composite numbers...

Knowing Our Numbers

While counting we use numbers to represent any quantity, to measure any distance or...

Join ALLEN!

(Session 2025 - 26)


Choose class
Choose your goal
Preferred Mode
Choose State
  • About
    • About us
    • Blog
    • News
    • MyExam EduBlogs
    • Privacy policy
    • Public notice
    • Careers
    • Dhoni Inspires NEET Aspirants
    • Dhoni Inspires JEE Aspirants
  • Help & Support
    • Refund policy
    • Transfer policy
    • Terms & Conditions
    • Contact us
  • Popular goals
    • NEET Coaching
    • JEE Coaching
    • 6th to 10th
  • Courses
    • Online Courses
    • Distance Learning
    • Online Test Series
    • International Olympiads Online Course
    • NEET Test Series
    • JEE Test Series
    • JEE Main Test Series
  • Centers
    • Kota
    • Bangalore
    • Indore
    • Delhi
    • More centres
  • Exam information
    • JEE Main
    • JEE Advanced
    • NEET UG
    • CBSE
    • NCERT Solutions
    • Olympiad
    • NEET 2025 Results
    • NEET 2025 Answer Key
    • JEE Advanced 2025 Answer Key
    • JEE Advanced Rank Predictor

ALLEN Career Institute Pvt. Ltd. © All Rights Reserved.

ISO