Subject: Compulsory Maths
Area is the quantity that expresses the extent of a twodimensional figure or shape in the plane.
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}
Find the length of a carpet of width 50cm to carpet a hall 8m long and 6m wide. Also, find the cost of carpeting at the rate of Rs 450/metre.
Solution:
Length of carpet(l_{1}) = ?
Breadth of carpet (b_{1}) = 50cm = 0.5m
Length of room (l) = 8m
Breadth of room (b) = 6m
While carpeting a room,
Area of carpet (A_{1}) = Area of room (A)
i.e. l_{1} * b_{1} = l*b
or, l_{1} = \(\frac{l*b}{b_{1}}\)
= \(\frac{8m*6m}{0.5m}\)
= \(\frac{48m^2}{0.5m}\)
= 96m
∴ 96m long carpet is required.
Again,
Length of carpet (l_{1}) = 96m
Cost of carpet/metre (C) = Rs 450/m
Toal cast (T) = l_{1} * C
= 96 * Rs 450
Rs 43,200
Hence, it costs Rs 43,200 in order to carpet the room.
Find the perimeter of the given window. Also, find its area, (\(\pi\) = \(\frac{22}{7}\))
Solution:
The given window frame consists of a rectangle and a semicircle.
For semicircle,
diameter (d) = 280cm
radius (r) = \(\frac{1}{2}\)d
=\(\frac{1}{2}\) * 280cm
= 140cm
Semicircle (c) =\(\frac{1}{2}\)\(\pi\)d
=\(\frac{1}{2}\)\(\frac{22}{7}\)280cm
=\(\frac{22 * 280}{14}\)cm
= 440cm
Now, perimeter of the window frame = l + b + b + semicircle
= 280cm + 150cm + 150cm + 440cm
= 1,020cm
Now, Area of rectangular part (A_{1}) = l * b
= 280cm * 150cm
= 42,000cm^{2}
Area of semi circle (A_{2}) =\(\frac{1}{2}\)\(\pi\)r^{2}
=\(\frac{1}{2}\)*\(\frac{22}{7}\)*(140cm)^{2}
= 30,800cm^{2}
∴ Area of window frame (A) = A_{1} + A_{2}
= 42,000cm^{2} + 30,800cm^{2}
= 72,800cm^{2}
Hence,
Perimeter of window frame (P) = 1020cm
Area ofwindow frame (A) = 72,800cm^{2} = 7.28m^{2}
Find the perimeter of the given plane figure.
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
A rectangular room is 8m long and 6m wide. What length of carpet of 3.5 m wide is needed to carpet room?
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.
What length of wall paper of breadth 50cm is reqiured to cover a wall of 5m long and 3m high?
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.
Find the number of stone to pave 12m long and 8m wide courtyard if the length of the stone is 2m and width is 1.6m?
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.
Find the area of the shaded region.
Solution:
From the figure,
Length of rectangle (l) = 50m
Breadth (b) = 30m
Width of the path (d) = 5m
Now,
Area of EFGH = l * d
= 50m * 5m
= 250m^{2}
Again,
Area of WXYZ = b * d
= 30m * 5m
= 150m^{2}
Also,
Area of crossing path ABCD = d^{2}
= (5m)^{2}
= 25m^{2}
Hence,
Area of shaded region = Area of EFGH + Area of WXYZ Area of crossing path ABCD
= 250m^{2} + 150m^{2} 25m^{2}
= 375m^{2}
Alternatively,
Area of crossing paths/shaded region = d(l + b  d)
= 5m (50m  30  5m)
= 5m * 15m
= 375m^{2}
The path having area 75m^{2} and width 2.5m is surrounded inside a garden. Find the length of the path.
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.
A room is 8m long, 6m wide and 4m height. It has door of 2m × 1.5m and two windows of size 2m × 3m. Find the
Solution:
Given,
Length of wall (l) = 8m
Breadth of wall (b) = 6m
Height of wall (h) = 4m
Length of door (l_{1}) = 2m
Height of door (h_{1}) = 1.5m
Length of window (l_{2}) = 2m
Height of window (h_{2}) = 3m
Cost of painting walls (C) = Rs 80/m^{2}
Now,
Area of 4 walls = 2h(l +b)
= 2 × 4m (8m + 6m)
= 8m × 14m
= 112m^{2}
Area of 4 walls and ceiling =2h(l +b) +lb
= 2 × 4m (8m + 6m) + 8m × 6m
= 8m × 14m + 48m^{2}
= 112m^{2} + 48m^{2}
= 160m^{2}
Area of 4 walls excluding door and windows (A) =2h(l +b) l_{1}h_{1} 2( l_{2}h_{2})
= 2 × 4m (8m + 6m) 2m × 1.5m  2(2m × 3m)
=8m × 14m  3m^{2}  2 × 6m^{2}
= 112m^{2}  3m^{2}  12m^{2}
= 97m^{2}
Cost of painting 4 walls and ceiling = C * A
= Rs 80/m^{2} × 97m^{2}
= Rs 7760
Hence,
Area of 4 walls = 112m^{2}
Area of 4 walls and ceiling = 160m^{2}
Area of 4 walls excluding door and windows = 97m^{2}
Cost of painting 4 walls and ceiling = Rs 7760
Calculate the volume and total surface area of the cylinder shown.
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) 
The cost of carpeting a room of 7m long, 6m wide at the rate of Rs.200/m is Rs.2400. Find the breadth of the carpet.
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.
Find the number of stones of size 40cm long and 30 cm wide to cover a courtyard of length 30 m and breadth is 15 m?
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.
Find the length of carpet of width 50cm to carpet a hall 8m long and 6 m wide. Also find the cost of carpeting at the rate of Rs.400/m.
Here,
Length of carpet (l_{1}) = ?
Breadth of carpet (b_{1}) = 50cm = 0.5m
Length of room (l) = 8m
Breadth of room (b) = 6m
While carpeting a room,
Area of carpet (A_{1}) = Area of room (A)
i.e. l_{1}*b_{1}= l*b
or, l_{1} = l*b / b_{1}
_{} = 8m * 6m / 0.5 m
= 96m
∴ 96m long carpet is required.
Again,
Length of carpet (l_{1}) = 96m
cost of carpet/ metre (C) = Rs.400/m
Total cost (T) = l_{1}*C
= 96 * Rs.400
= Rs.38400
Hence, it costs Rs.38400 in order to carpet the room.
A hall is 30 m long and 20 wide. The area of four wall is 400 m^{2}. Find the height of the wall. Also, find
a) The area of four wall and ceiling.
b) The area of the four wall, ceiling and floor (T.S.A).
c) The area of floor and ceiling.
d) Find the cost of painting four wall and ceiling at the rate of Rs.200/m^{2}
Hence,
length (l) = 40m
breadth (b) = 30 m
area of 4 walls (A) = 600m^{2}
height (h) =?
Area of 4 walls (A) = 400m^{2}
i.e. 2h(l + b) = 400 m^{2}
or, 2h( 30m + 20m) = 400m^{2}
or, 2h * 50m = 400m^{2}
or, h = 4m
∴ height of the wall (h) = 4m
a) Now, Area of 4 walls and ceiling
= 2h(l + b) + lb
= 2*4m( 30m + 20m) + 30m*20m
= 400m^{2} + 600m^{2}
= 1000m^{2}
b) Area of 4 walls, ceiling and floor (T.S.A)
= 2h (l + b) + 2lb
= 2*4m (30m + 20m ) + 2*30m*20m
= 400m^{2} + 1200m^{2}
= 1600m^{2}
c) Area of floor and ceiling
= 2 lb
= 2*30m*20m
=1200m^{2}
d) Area of 4 walls and ceiling (A) = 1000m^{2}
cost of painting the wall (C) = Rs.200/m^{2}
Total cost (T) = C*A
= Rs.200/m^{2}*1000m^{2}
= Rs.200000
Total cost of painting four wall and ceiling is Rs.200000.
A room is 5m long, 4m wide and 3m high. Find the following:
a) Area of floor of a room
b) Area of 4 walls of a room
c) Area of ceiling
d) Area of 4 wall and ceiling
e) Area of 4 wall, ceiling and floor (TSA of a room)
Here,
length (l) = 5m
breadth (b) = 4 m
height (h) = 3m
Now,
a) Area of floor (A)
= l*b
= 5m * 4m
= 20m^{2}
b) Area of 4 walls (A)
= 2h (l + b)
= 2*3m (5m + 4m)
= 54m^{2}
c) Area of ceiling (A)
= l * b
= 5m * 4m
= 20m^{2}
d) Area of 4 wall and ceiling (A)
= 2h (l + b) + lb
= 2*3m (5m + 4m) + 5m * 4m
= 54 m^{2}+ 20m^{2}
= 74 m^{2}
e) Area of 4 wall, ceiling and floor (TSA)
= 2h (l + b) + lb + lb
= 2*3m(5m + 4m) + 5m * 4m + 5m * 4m
=54m^{2} + 20m^{2}+ 20m^{2}
=94m^{2}
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