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Figure possess certain length, breadth, height, perimeter, area, volume, etc. The area is the quantity that expresses the extent of a twodimensional figure or shape, or planar lamina, in the plane. And mensuration is the branch of mathematics which deals with the study of Geometric shapes, their area, volume and related concepts. This chapter will you to apply those expressions into practical life.
There are different formulas to calculate different shapes like land, playground, water tank, circular track, etc.
The formula for different shape to find area, perimeter or circumference are given below:
Name of Figure  Diagram  Area  Perimeter/circumference 
Square  A=l^{2}sq. unit  P = 4l unit  
Rectangle 
A=...... sq. unit  P = 2(l=b) unit  
Rhombus  A= ....sq. unit  P = ...... unit  
Parallelogram  A= ....sq. unit  P = ...... unit  
Triangle  A= ....sq. unit  P =...... unit  
Equilateral triangle  A= ....sq. unit  P =...... unit  
Quadrilateral  A= ....sq. unit  P = sum of four sides  
Trapezium  A= ....sq. unit  P = sum of four sides  
Circle  A= ....sq. unit  c = 2πr =πd 
Areas of pathways outside a rectangle  Areas of pathways inside a rectangle 
Let 'd' be the width of the pathway running uniformly outside a rectangle ABCD. The outer rectangle EFGH will be formed whose length and breadth would be (l+2d) and (b+2d) respectively. Now, area of ABCD = l * b Area of EFGH = (l + 2d).(b + 2d) \(\therefore\) Area of path =Area of EFGH  area of ABCD = (l + 2d).(b + 2d) l * b = (lb + 2ld + 2bd + 4d^{2}  lb) = 2ld + 2bd +4d^{2} = 2d(l + b + 2d) So, area of pathway running outside a rectangle =2d(l + b + 2d)
For a square shape, l = b \(\therefore\) Area of pathway running outside square = 2d(l + l + 2d) = 2d(2l + 2d) = 2d.2(l + d) = 4d(l + d) 
Let 'd' be the width of the uniform pathway running inside a rectangle ABCD having length l and breadth b. The inner rectangle EFGH would have length and breadth (l  2d) and (b  2d) respectively. Now, area of ABCD = l * b area of EFGH = (l  2d) * (b  2d) \(\therefore\) Area of path =area of ABCD  area of EFGH = l * b (l  2d) * (b  2d) = lb  (lb  2id  2bd + 4d^{2}) = lb  lb + 2ld + 2bd 4d^{2} = 2ld + 2bd 4d^{2} = 2d(l + b  2d) So, area of pathway running inside a rectangle =2d(l + b  2d)
For a square shape, l = b \(\therefore\) Area of pathway inside a square = 2d(l + l  2d) = 2d(2l  2d) = 2d.2(ld) = 4d(l  d) 
Method 1  Method 2 
Let 'd' be the width of the crossing pathways. Area of path EFGH = l * b Area of path WXYZ = b * d Area of crossing path PQRS = d^{2} [\(\therefore\) PQRS is a square shaped] The area of crossing paths = Area of (EFGH + WXYZ  PQRS) = (ld + bd  d^{2}) = d(l + b  d)

Let 'd' be the width of crossing pathways. The area of crossing pathways can be obtained by subtracting the area of APQT from the area of ABCD. \(\therefore\) Area of crossing paths = Area of (ABCD  APQT) = (l * b) (l  d)(b  d) = lb  (lb  ld bd + d^{2}) = lb  lb + ld + bd  d^{2} = ld + bd  d^{2} = d(l + b  d)

Hence, the area of crossing pathways = d(l + b  d)
For square field, l = b
\(\therefore\) Area of crossing paths = d(l + l  d) = d(2l  d)
Outside a circular field  Inside a circular field 
Let 'd' be the width of the circular pathway around a circular field of radius r. Area of ABC = \(\pi\)r^{2} Area of PQR =\(\pi\)(r + d)^{2} \(\therefore\) Area of circular path = Area of ( PQR  ABC) = \(\pi\)(r + d)^{2}\(\pi\)r^{2} 
Let 'd' be the width of the circular pathway inside a circular field of radius r. Area of ABC = \(\pi\)r^{2} Area of PQR =\(\pi\)(r + d)^{2} \(\therefore\) Area of circular path = Area of (ABC  PQR) 
Eventually, we relate area to cost. Cost is estimated to cover the pathways with the help of bricks, tiles or stones. Cost estimation is based on area and rate of a unit square.
Let,
A = area of pathways
R = rate of unit square
T = total cost
Then, T = A * R
If N = the number of bricks (tiles or stones) required to pave. and
a = surface area of a single brick.
Then, N =\(\frac{A}{a}\)
Here we learn how to calculate the cost of plastering and painting the walls of a room.
Look at the following figures and discuss.
In the object, we can see 6 faces. Face 1 as a floor, face 2,3,4 and 5 as walls and face 6 as a ceiling. It is easy to calculate areas with this figure.
Here, Area of floor = l * b
Area of ceiling = l * b
Area of right wall = b* h
Area of left wall = b * h
Area of front wall = l * h
Area of back wall = l * h
Now, Area of 4 walls = Area of (left wall + right wall + front wall + back wall)
= bh + bh + lh + lh
= 2bh + 2lh
= 2h (l + b)
Area of 4 walls, fool and ceiling = 2h(l + b) + lb + lb
= 2h(l + b) + 2lb
= 2lh + 2bh + 2lb
= 2(lh + bh + lb) For square base room,
Length (l) and breadth (b) are equal but height (h) is difference
Thus, Area of floor = l * b
= l * l
= l^{2}
Area of ceiling = l * b
= l * l
= l^{2}
Area of 4 walls = 2h(l + b)
= 2h(l + l)
= 2h * 2l
= 4hl
Total surface area = 2(lb + bh + lh)
= 2(ll + lh +lh)
= 2(l^{2} + 2lh)
= 2l(l + 2h)
For cubical room,
Here lenght (l), breadth (b) and height (h) are equal i.e. l = b = h = a
Thus, Area of floor = l * b = a * a = a^{2}
Area of ceiling = l * b = a * a = a^{2}
Area of 4 walls = 2h(l + b)
= 2a(a + a)
= 2a*2a
= 4a^{2}
Total surface area = 2(lb + bh + lh)
= 2(a.a + a.a + a.a0
= 2(a^{2} +a^{2} +a^{2})
= 2 * 3a^{2}
= 6a^{2}
Simply, all the faces of the cubical room are square in shape.
Area of one face = a^{2}
Area of walls = 4a^{2}
Total surface area = Area of walls = 6a^{2}
In the presence of door or window,
Suppose, a room contains a door having length l_{1}, and height h_{1} and a window having length l_{2} and height h_{2}.
In order to calculate the area of 4 walls, the area of door and window should be subtracted from the area of 4 walls and window should be subtracted from the area of 4 walls.
Here, Area of 4 walls = 2h(l + b)
Area of 4 walls excluding door = 2h(l + b)  l_{1}h_{1}
Area of 4 walls excluding a door and a window = 2h(l +b)  l_{1}h_{1}  l_{2}h_{2}
Solution:
Length (l) = 7.5cm
Breadth (b) = 5cm
Now,
Area of given rectangle = l * b
= 7.5cm * 5cm
= 37.5cm^{2}
Also,
Perimeter = 2(l + b)
= 2 (7.5 + 5)cm
= 2 * 12.5cm
= 25cm
Solution:
Length of carpet (l_{1}) = ?
Breadth of carpet (b_{1}) = 3.5m
Length of room (l) = 8m
Breadth of room (b) = 6m
Area of carpet (A_{1}) = Area of room (A)
i.e. l_{1} * b_{1} = l * b
or, l_{1} = \(\frac{l*b}{b_{1}}\)
= \(\frac{8m*6m}{3.5}\)
= \(\frac{48m^2}{3.5m}\)
= 13.71m
∴ 13.71m long carpet is required.
Solution:
Length of wall paper (l_{1}) = ?
Breadth of wall paper (b_{1}) = 50cm = 0.5m
Length of wall (l) =5m
Breadth of wall (b) =3m
While carpeting a room,
Area of wall paper (A_{1}) = Area of wall (A)
i.e. l_{1} * b_{1} = l*b
or, l_{1} = \(\frac{l*b}{b_{1}}\)
= \(\frac{5m*3m}{0.5m}\)
= \(\frac{15m^2}{0.5m}\)
= 30m
∴ 30m long wall paper is required.
Solution:
Length of courtyard (l) = 12m
Breadth of courtyard(b) = 8m
Area of courtyard (A) = l * b
= 12m * 8m
= 96m^{2}
Again,
Length of stone (l_{1}) = 2m
Breadth of stone (b_{1}) = 1.6m
Area of stone (a) = l_{1} * b_{1}
= 2m * 1.6m
= 3.2m^{2}
Now,
Required number of stone(N) = \(\frac{Area of courtyard}{Area of stone}\)
= \(\frac{96m^2}{3.2m^2}\)
= 30
Hence, 30 stones are needed to pave the courtyard.
Solution:
Area of square garden (A) = 75m^{2}
Breadth (b) = 2.5m
Length (l) = ?
Now,
Area of square = l * b
or, 75m^{2} = l * 2.5m
or, l = \(\frac{75m^2}{2.5m}\)
∴ l = 30m
Hence, the length of the path is 30m.
Volume  =  π r²h = π × 4² × 6 = 96 π 
=  301.5928947 cm³  
=  302 cm³ (to 3 significant figures) 
Area of curved surface  =  2π rh = 2 × π × 4 × 6 
=  48π  
=  150.7964474 cm² 
Area of each end  =  π r² = π × 4² 
=  16π  
=  50.26548246 cm² 
Total surface area  =  150.7964474 + (2 × 50.26548246) 
=  251.3274123 cm²  
=  251 cm² (to 3 significant figures) 
Here,
Length of the room (l) = 7m
Breadth of room (b0 = 6m
Cost of carpeting the room (C) = Rs.200/m
Total cost (T= Rs.2600
∴ Length of carpet (l_{1}) = T/C
= Rs.2400/Rs.200 = 12m
Breadth of carpet (b_{1}) = ?
When we carpet the room,
Area of carpet = Area of room
i.e. l_{1}*b_{1} = l*b
or, 12m*b_{1} = 7m*6m
or, b_{1} = 42m^{2}/12m
or, b_{1} = 3.5 m
∴ The breadth of the carpet is 3.5m.
Here,
Length of stone (l_{1}) = 40 cm = 0.4m
Breadth of stone (b_{1}) = 30 cm = 0.3m
∴ Area of each stone (a) = l_{1}*b_{1}
= 0.4m * 0.3m = 0.12m^{2}
And, length of courtyard (l) = 30m
breadth of courtayard (b) = 15m
∴ Area of courtyard (A) = l*b
= 30 m * 15 m
= 450 m^{2}
Now,
Required number of stones (N) = Area of courtyard(A) / Area of stone(a)
= 450m^{2} / 0.12m^{2}
= 3750
Hence, 3750 stones of given dimension are required to cover the courtyard.
In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A hall is 14 m long and 11 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be: 
Find the surface area of a 10cm*4cm*2cm brick.
A boat having a length 4 m and breadth 3 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:
The curved surface area of a cylindrical pillar is 264 m^{2} and its volume is 924 m^{3}. Find the ratio of its diameter to its height.
How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?
A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?
A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:
A circular well with a diameter of 2 meters, is dug to a depth of 14 meters. What is the volume of the earth dug out.
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what will be the cost of plastering the four walls of a room whose perimeter is36m and height3m,at Rs.16 per square metre?
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A Wall is 25m long 4m high and 30cm thick.If it contains two windows of 2m by 1.5m each and a door of 3m by 2m,Find the number of bricks of size 15cm*10cm*5cm to build the wall.Also find the total cost of bricks at the rate of Rs8.50 per brick.
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