Mensuration and its Applications By – Aparup Dey 10 ‘D’ Roll# 10

CONTENTS• Triangle

• Square• Rectangle• Parallelogram• Trapezium• Circle• Cuboid

• Cube• Cylinder• Cone• Sphere• Applications Of

Mensuration

** Click to jump directly to figure

TRIANGLE• A Triangle is a three sided rectilinear figure•Sum of interior angles = 180°

FORMULAE•Perimeter = a+b+c• Area = ½ .base.height

Heron’s Formula•If ABC is a triangle with sides a,b,c then•Area = √{s(s-a)(s-b)(s-c)} where s is the semi perimeter• s = (a+b+c)/2

Heron’s Formula : Derivation

Square• A square is a simple rectilinear figure in which

all sides are equal• Each interior angle is equal to 90°

FORMULAE• Perimeter= a+a+a+a

= 4a• Area = a*a

= a2 • (Diagonal)2 = a2+a2 (Pythagoras theorem )

d= √(2a2)

d= √2 a

RECTANGLE• A rectangle is a simple rectilinear 4 sided figure in which

opposite sides are equal in length.• Each interior angle is equal to 90°.

FORMULAE• Perimeter = l+l+b+b

= 2(l+b) • Area = l*b• (Diagonal)2 = l2 + b2 (Pythagoras Theorem)

d = √(l2+b2)

Parallelogram• A parallelogram has opposite sides parallel and equal in

length. • Also opposite angles are equal.

FORMULAE• BFEC is a rectangle• Area = l*b = h*BC• Now triangle DEC is removed and CE is placed along h• The resulting figure is a parallelogram• The area will remain same• Area = Height * Corresponding Base ** here A = h.BC or h.AD

TRAPEZIUM• A trapezium has a pair of opposite sides parallel.• It is called an Isosceles trapezoid if the sides that aren't

parallel are equal in length and both angles coming from a parallel side are equal, as shown.

FORMULAE• Perimeter = AB+BC+CD+DA• Area = area of triangle ABD+BCD• = ½ AB.DF + ½ DC.BE *let DF=BE=h

• = ½ (AB.h + DC.h) since DFBE is a rectangle

• = ½ (AB+DC)h • = ½ (sum of parallel sides).Height

CIRCLE• A circle is the set of all those points, say P, in a plane, each which is at a

constant distance from a fixed point in that plane.

FORMULAE(r is the radius)

1. Circumference and area of a circle• Cirumference = 2πr• Area = πr2

2. Area of a circular ring

(R and r be the radii of the bigger and smaller circles)

Area = πR2 – πr2

= π(R2-r2)

3. Perimeter and area of semicircle

P = ½ .2πr+2r = (π+2)r

A= ½ πr2

Cuboid• A solid in the shape of a box (or brick) is called a cuboid.• It has 6 rectangular plane surfaces called faces.

FORMULAE• Surface Area = Sum of areas of 6 faces

=lb+lb+bh+bh+hl+hl

=2(lb + bh + hl )• Lateral surface area = sum of areas of 4 walls

= lh + lh + bh + bh

= 2( l+b )h• Volume = area enclosed by the cuboid = lbh• Length of diagonal = √(l2+b2+h2) ** the base diagonal is l√(l2+b2) and height is h

cube• A rectangular solid bounded by 6 squares is called a

cube.• It is a cuboid in which l = b = h = a(assumption).

FORMULAE• Surface area = 6a2

• Lateral Surface Area = 4a2

• Volume = a3

• Length of Diagonal = √(3a2) = √3 a

cylinder• A solid obtained by revolving a rectangular lamina about

one of its sides is called a right circular cylinder.

FORMULAE1.) Solid Cylinder• Curved Surface Area = 2πrh **area of rectangular lamina ; l=2πr & b= h

• Total surface Area = Curved S.A+

Area of 2 circular ends

= 2πrh+2.πr2

= 2πr(h + r)• Volume = Area of cross section * Height

= πr2h

cylinder2.) Hollow Cylinder• Thickness = R-r• Area of cross-section = πR2-πr2

= π(R2-r2)

= π(R+r)(R-r)• External Curved Surface Area = 2πRh• Internal Curved Surface Area = 2πrh• Total Surface Area = External C.S.A+ Internal C.S.A

+ Area of 2 ends

= 2πRh + 2πrh +2π(R2-r2)

= 2π(Rh+rh+R2-r2)• Volume of material = πR2h- πr2h

= π(R+r)(R-r)h

Cone• A solid obtained by revolving a right angled triangular

lamina about any side (other than the hypotenuse) is called a right circular cone.

FOUMULAE• Slant height (s) = √(r2+h2) ** Pythagoras Theorem

• Curved surface Area = πrs• Total Surface Area = C.S.A+Area of Base

= πrs+πr2

= πr(s+r)• Volume = 1/3πr2h

Derivation of Lateral Surface Area of a Cone

• s = radius of circle of which the cone is a part• So area of the Larger circle = πs2

• Circumference = 2πs• The arc AB originally wrapped around the base of the

cone, and so its length is the circumference of the base. Recall that circumference of a circle is given by 2πr

• The ratio of area x of the shaded sector to the area of the whole circle, is the same as the ratio of the arc AB to circumference of the whole circle*. Put as an equation

x/πs2 = 2πr/2πs

x = πrs

sphere• A solid obtained by revolving a circular lamina about any

of its diameters is called a sphere.

FORMULAE1.) Solid Sphere • Surface Area = 4πr2

• Volume = 4/3πr3

2.) Spherical shell• Thickness = R-r• Volume = 4/3πR3-4/3πr3

= 4/3π(R3-r3)

sphere3.)Hemisphere• Curved Surface Area = ½ Surface area of Sphere• = ½ . 4πr2 = 2πr2

• Total Surface Area = Curved S.A.+ area of base = 2πr2+πr2 = 3πr2

• Volume = ½ (volume of Sphere) = ½ * 4/3πr3 = 2/3πr3

4.) Hemispherical Shell• Thickness = R-r• Area of Base = π(R2-r2) = π(R+r)(R-r) • External Curved S.A = 2πR2

• Internal Curved S.A = 2πr2

• Total S.A. = 2πR2-2πr2+π(R2-r2) = π(3R2+r2)• Volume = 2/3π(R3-r3)

Applications of Mensuration• Finding area of a plot of land being bought or sold• Finding area of a parking lot to be paved.• Finding the Length of boundary to be fenced.• In cooking, we use tools such as measuring jugs may be used to determine

volumes.• When planning a car journey, we may look at a map to find out the quickest

way to reach a particular distance.• Before leaving the door, we may check the weather forecast. The volume of

mercury in thermometers will indicate the weather.• To find the volume of material required to make something. For eg –

trophies and showpieces• If we want to wrap or cover something, then we can find the exact area of

paper or cloth required by finding the surface area of the solid.• To find the cost of painting the walls.• To make ice-cream cones and pencils, etc.