# Mensuration-Notes

Class VIII Math
Notes for Mensuration
STANDARD UMTS OF VOLUME
1 dm = 10 cm
1 dm3 or litre = 1 dm × 1 dm × 1 dm = (10 × 10 × 10) cm3 = 1000 cm3
1 cm = 10 mm
1m =100 cm
1m3 = (100 × 100 × 100) cm3 = 1000000 cm3
1 m3 = 1000 × 1000 cm3 = 1000 litre
1 kilolitre = 1m3 = 1000 litre
(mm) millimeter, ( cm) centimetre, (dm) decimetre, (m) metre
Plans figures
The geometrical figures which have only two dimensions are called as the plane figures.
Booster 1
A square with sides of 1 cm has an area of 1 cm2.
Find the area of the shaded shape.
Explanation
The shape covers 11 squares, so its area is 11 cm2.
Booster 2
Find the area of the shaded triangle.
Explanation
The triangle covers 6 full squares marked F, and 4 half squares marked H. Area = 6 + 2 = 8 cm2.
Booster 3
Estimate the area of the shape shaded in the diagram.
Explanation
This is a much more complicated problem as there are only 9 full squares marked F, but many other part square. You need to combine part squares that approximately make a whole square. For example, the squares marked make about 1 full square; the squares marked � make about 1 full square; the squares marked + make about 1 full square; the squares marked • make about 1 full square. Thus the total area is approximately
9 + 4 = 13 cm2.
Triangle
(i)
Area of triangle
(ii) Area of an equilateral triangle
(iii) Area of an isosceles triangle :- base = b, equal side = a

Booster 4
The parallel sides of a trapezium are 20 cm 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 || AB.
∴ ALCD is a parallogram.
⇒ AL = CD = 10 cm and CL = AD = 13 cm
In ΔCLB, we have CL = CB = 13 cm
∴ΔCLB is an isosceles triangle.
[∵BL= AB – AL = (20 – 10) cm = 10 cm]
Applying pythagoras theorem in δCML, we have
CL2 = CM2 + LM2
132 = CM2 + 52
CM2 = 169 – 25 = 144
Area of parallelogram ALCD = AL × CM = (10 × 12) = 120 cm2
Hence, Area of trapezium ABCD = Area of parallelogram ALCD + Area of δCLB = (120 + 60) cm2 = 180 cm2
Booster 5
If the area of a rhombus be 24 cm2 and one of its diagonals be 4 cm, 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 O.
Area of rhombus ABCD = 24 cm2
Thus, we have AC = 4 cm and BD = 12 cm
Since the diagonals of a rhombus bisect each other at right angle. Therefore, δOAB is right triangle, right angled at 0.
Using pythagoras theorem in δOAB, we have
AB2 = OA2 + OB2
AB2 = 22 + 62 = 40

Hence, perimeter of rhombus ABCD
Booster 6
Find the area of a regular octagon each of whose sides measures 4 cm.
Explanation
Area of the octagon
Booster 7
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 = 20m3.
Cube
A cuboid whose length, breadth and height are all equal is called a cube.
(i)   Surface area of a cube . Since alI the faces of a cube are squares of the same size i.e., for a cube we have l = b = h. Thus if l cm 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
(ii)  Lateral surface area of the cube = 2(l × l + l × l)
= 2(l2 + l2) = 4l2 = 4(Fdge)2
(iii)  Volume of a cube = l × l × l = l3
Booster 8
Find the total surface of a hollow cylinder open at ends, if the length is 12 cm, the external diameter 10 cm and thickness 2 cm.
Explanation
The outer radius (R) = 5 cm, Thickness = 2 cm
∴. Inner radius (r) = 5 cm – 2 cm = 3 cm
Outer curved surface of the cylinder
= 2πrh = 2 × π × 5 × 12 = 120πcm2
Inner curved surface of the cylinder 27πrh
= 2 × π × 3 × 12 = 72 π cm2.
Both ends of the cylinder will be of the shape, as shown in figure (ii).
∴. Area of one of end of the cylinder = πR2 – πr2
= π × 52 – π × 32 = 25π – 9π = 16π cm2.
Area of both ends = (2 × 16π) cm2 = 32π cm2
∴Total surface of the cylinder = External curved surface + Internal curved surf ace + 2 (Area of the base of the ring)

Remark : It is advisible to put the value of π is the end in such calculations.
Circular cone
There are many objects around us which are conical in shape like an ice-cream cone, a conical, tent a birthday cap etc. These objects are of the shape of a right circular cone.
δAOC is right angled at O.
By Pythagoras theorem,
AC2 = AO2 + OC2
i.e. l2 = h2 + r2

(ii)  Total surface area of cone = Curved surface area of cone + Area of the circular base
= πrl + πr2
= πr (r + l) sq. units.
(iii) Volume of a cone (where ‘r’ r is the radius and ‘h’ is the height of cone).
Sphere and hemisphere
Sphere: The set of all the points in space which are equidistant from a fixed point is called a sphere.
Hemisphere: A lane through the centre of a sphere divides the sphere into two equal parts and each part is called a hemisphere.
(i)   Curved surface area of hemisphere = 2πr2
(ii)  Total surface area of hemisphere 2πr2 + πr2 = 3πr2
Booster 9
If radius of the base of a cone is 140 dm and its slant height is 9 m. Find the
(i) curved surface area
(ii) total surface area
Explanation
Radius of the base of the cone (r) = 140 dm = 14 m
Slant height (l) = 9 m
(i)   Curved surf ace area of the cone
(ii)  Total surface area of cone = πrl + πr2

Question 12
The surface area of a sphere is 2464 dm2. Find its diameter.
Solution
Surface area of a sphere = 2464 dm2
Therefore, 4πr2 = 2464

Diameter of the given sphere = 2 × 14 = 28 dm.
Booster 10
The dome of a building is in the form of a hemisphere of radius 63 dm.
Find the cost if it is to be painted at the rate of Rs 5 per m2.
Explanation
Radius of the hemisphere = 63 dm
Surface area of hemisphere = 2πr2

Booster 11
The earth taken out while digging a pit, is evenly sprea 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 × 60m × h
height of field raised = 0.57 m