AREAS AND VOLUMES OF SOLID STRUCTURES

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AREAS AND VOLUMES OF SOLID STRUCTURES

AREAS AND VOLUMES OF SOLID STRUCTURES

TYPE - II

AREAS AND VOLUMES OF SOLID STRUCTURES

1.CYLINDER

*Curved surface area = 2?rh
*Total surface area
= 27?r2 + 27?rh = 27?r(r+h)
*Volume = ?r2h

EXAMPLES

1.
If the curved surface area of a cylinder is 450 cm and the radius is 7 cm, find the total surface area.
Curved surface area = 2?rrh = 450cm
Total surface area = 27?r2 + 2?rh
= 2? x 7 x 7 + 450
= 2x22/7x7x7+450
=308 + 450=758cm2
2.
If the height and diameter of a Cylinder
are 49cm and 20 cm respectively, find its volume.
V =?r2h r= 20/2
=22/7*49*l0*l0
= 15400
.'.Volume of the cylinder = 15400cm3

2.CUBE

*Volume, V = (side)3
=a3
*Total Surface Area,A=6a2
*Diagonal, d = -/3a
*Area of 4 walls, A=4a2

EXAMPLES

1.
Find the surface area of a cube whose volume is 343m3
Volume = a3 = 343m3
.'.a =3 /343 = 7m
.'.Surface area= 6a2
=6x7x7 = 294m2
2.
How many bricks will be needed to construct a wall 4m long, 13cm thick and 3m high if each brick measures 20cmx 12cmx65cm
Volume of wall = 4 x 3 x 0.13m
Volume of 1 brick = 0.2 x 0.12 x 0.065m
Required no. of bricks =4x3x0.13/0.2*0.12*0.065
=1.56/ .0,0156
=1.56x100000 / .00156x100000
= 1000
.'.Required no. of bricks= 1000
*A cube of sides 8cm is melted and smaller cubes of sides 1cm each are formed. How many such cubes are possible
Required number=(Original length of side/New length of side)
=(8/1)3 =512
Volume of cuboid,
V= Ibh
Total surface area,
A = 2(lb+bh+lh)
Area of 4 walls of a cuboidal room,
A = 2h(l+b)
Diagonal, d = l2+b2+h2

EXAMPLES

1.
Find the volume of a cuboid having 0.03m length, 500cm breadth and 760mm height.
Volume = Ibh
= 0.03mx500cmx760mm
= 0.03m x 5m x 0.76m
= 0.1140m3
2.
Find the maximum possible length of rod that can be placed inside a box of length 15cm, breadth 9cm and height 13cm.
The maximum possible length inside the box will be length of its diagonal
d = 1^2+b^2+h^2
= l5^2+9^2+13^2
= 225 +81 + 169 = /475
= 5 /19
.'.Required length = 5 V19 cm

4.CONE

Volume=1/3?r2h =
1/3* base area * height
*Curved surface area,
A = ?rl, where 'r' is the radius of the base and 'I' is the slant height
*Slant height, I = /h2 +r2

EXAMPLES

1.
The ratio of radius and height of a cone is 5:12. If its volume is 8478cm3, find its slant height.
Let 'r' be 5x and 'h' be 12x
Volume =1/3?r2 h=8478
1/3*22/7*25x2*12x=8478
x3=8478x3x7/25x12x22
x3 = 27
.'.x=3
.'.r = 15cm and h = 36 cm
Now, slant height = /r2 +h2
= /l5^2 + 36^2
= /225+1296
= 1521 =39
.'.Slant height = 39 cm
2.
Find the radius of the base of a cone, with volume 1600cm2 and radius 5cm.
Volume = 1/3?r2h
1600 = 1/3x22/7x5x5xh
h=1600x3x7/22x5x5 = 61.09
.'.height = 61.09cm

5.SPHERE

*Volume =4/3 ?r3
*Curved surface area = 47?r2

6.HEMISPHERE

*Volume = 2/3 T?r3
*Curved surface area = 2?r2
*Total surface area = 3?r2

*VOLUME OF SPHERICAL SHELL

=4/3?(R3-r3)
*VOLUME OF A METAL IN A HOLLOW PIPE = ?h(R2- r2)
*TOTAL SURFACE AREA OF AN OPEN PIPE = 2?[Rh + rh + (R2 - r2),
Where R is external radius

EXAMPLES

1.
If the radius of a sphere is tripled, the volume increases by:
When radius is r, volume V1 = 4/3?r3
When radius is tripled, ie 3r, volume
v2 = 4/3 ?(3r)3
= 4/3? 27r3 = 27x 4/3 ?r3
.'.27 times initial volume
ie, V2 = 27V1
.'.Increase in volume =27v1-v1/v1*100
=26V/V1*100
= 2600%
2.
The volume of a hemisphere of radius 21cm is___
V= 2/3?r3
= 2/3x22/7-x 21x21x21
=19404cm3
3.
Find the surface area of a sphere of radius 63cm.
Surface area = 47?r2
=4x22/7x63x63
=49,896cm2

PRACTICE EXERCISE

1.
The length and breadth of a room is in the ratio of 3:2. If its perimeter is 80m, then its length will be
2.
How many bricks of 15cm x 10cm will be required to pave the floor of a hall which is 30m long and 20m wide
3.
Ajith took 15 seconds to cross a rectangular field diagonally walking at the rate of 52m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. Find the area of the field
4.
A rectangular lawn 80 metres by 50 metres has two roads each 5 metres wide, running in the middle of it, one parallel to length and the other parallel to the breadth. Find the cost of gravelling them at 80 paise per m2
5.
If the radius of a cylinder is halved and the height is doubled, what is the ratio between the new volume and previous volume.
6.
2m3 of gold sheet is extended by hammering 80 as to cover an area of 1 hectare. The thickness of the gold sheet is___
7.
The area of a square plot is twice the area of a rectangular plot. If the length of the rectangular plot is 32m and the area of the square plot is 2560cm2, what is the breadth of the rectangle
8.
Two adjacent sides of a parallelogram are 4cm and 5 cm. If the diagonal is 7cm, find the area
9.
The perimeter of an isosceles triangle is 150cm. If the length of equal sides is 30cm each, find the length of the base.
10.
A rectangular hall 10m long 8m broad is surrounded by a verandah 2 metres wide, Find the area of the Verandah.

ANSWERS WITH EXPLANATIONS

1.
Let 3x be the length and 2x be the breadth
Then, 2(3x + 2x) = 80
10x = 80
.'.x=8
.'.Length = 3x = 24m

2.
No. of bricks =Area of floor/Area of a brick
=2000cm x 3000cm/15cm x 10cm
= 40,000
3.
Length of diagonal (52*15/60)m
= 13m
Length + breadth = (68*15/60) m
= 17m
.'./l^2+b^2 = 13
ie, 1^2 + b^2= 169 ----(i)
l + b = 17-------(ii)
Solving (i) and (ii) =>
Area = lb = 1/2 (21b)
=1/2[(1+b)2-(1^2+b^2)]
= 1/2[(17)^2-169]
= 1/2[289-169] =60m2
.'.Area of the field = 60m2
4.
Area of the road ABCD = 80 x 5 = 400m2 Area of the road EFGH=50*5=250m2

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