Notes on Area | Grade 9 > Compulsory Maths > Mensuration | KULLABS.COM

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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 two-dimensional 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=l2sq. 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

 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 + 4d2 - lb) = 2ld + 2bd +4d2 = 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 + 4d2) = lb - lb + 2ld + 2bd -4d2 = 2ld + 2bd -4d2 = 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(l-d) = 4d(l - d)

Area of pathways crossing each other perpendicularly

 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 = d2 [$$\therefore$$ PQRS is a square shaped] The area of crossing paths = Area of (EFGH + WXYZ - PQRS) = (ld + bd - d2) = 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 + d2) = lb - lb + ld + bd - d2 = ld + bd - d2 = 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)

Area of path around a circular field

 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 = $$\pi$$r2 Area of PQR =$$\pi$$(r + d)2 $$\therefore$$ Area of circular path = Area of ( PQR - ABC) = $$\pi$$(r + d)2-$$\pi$$r2 Let 'd' be the width of the circular pathway inside a circular field of radius r. Area of ABC = $$\pi$$r2 Area of PQR =$$\pi$$(r + d)2 $$\therefore$$ Area of circular path = Area of (ABC - PQR)

Relation between area, cost and quantities

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 =$$\frac{A}{a}$$

Area of 4 walls, floor and ceiling

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

= l2

Area of ceiling = l * b

= l * l

= l2

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(l2 + 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 = a2

Area of ceiling = l * b = a * a = a2

Area of 4 walls = 2h(l + b)

= 2a(a + a)

= 2a*2a

= 4a2

Total surface area = 2(lb + bh + lh)

= 2(a.a + a.a + a.a0

= 2(a2 +a2 +a2)

= 2 * 3a2

= 6a2

Simply, all the faces of the cubical room are square in shape.

Area of one face = a2

Area of walls = 4a2

Total surface area = Area of walls = 6a2

In the presence of door or window,

Suppose, a room contains a door having length l1, and height h1 and a window having length l2 and height h2.

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) - l1h1

Area of 4 walls excluding a door and a window = 2h(l +b) - l1h1 - l2h2

• There are different formulas to calculate different shapes like land, playground, water tank, circular track, etc.
.

Very Short Questions

Solution:

Length (l) = 7.5cm

Breadth (b) = 5cm

Now,

Area of given rectangle = l * b

= 7.5cm * 5cm

= 37.5cm2

Also,

Perimeter = 2(l + b)

= 2 (7.5 + 5)cm

= 2 * 12.5cm

= 25cm

Solution:

Length of carpet (l1) = ?

Breadth of carpet (b1) = 3.5m

Length of room (l) = 8m

Breadth of room (b) = 6m

Area of carpet (A1) = Area of room (A)

i.e. l1 * b1 = l * b

or, l1 = $$\frac{l*b}{b1}$$

= $$\frac{8m*6m}{3.5}$$

= $$\frac{48m^2}{3.5m}$$

= 13.71m

∴ 13.71m long carpet is required.

Solution:

Length of wall paper (l1) = ?

Breadth of wall paper (b1) = 50cm = 0.5m

Length of wall (l) =5m

Breadth of wall (b) =3m

While carpeting a room,

Area of wall paper (A1) = Area of wall (A)

i.e. l1 * b1 = l*b

or, l1 = $$\frac{l*b}{b1}$$

= $$\frac{5m*3m}{0.5m}$$

= $$\frac{15m^2}{0.5m}$$

= 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

= 96m2

Again,

Length of stone (l1) = 2m

Breadth of stone (b1) = 1.6m

Area of stone (a) = l1 * b1

= 2m * 1.6m

= 3.2m2

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) = 75m2

Breadth (b) = 2.5m

Length (l) = ?

Now,

Area of square = l * b

or, 75m2 = 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 (l1) = T/C

= Rs.2400/Rs.200 = 12m

Breadth of carpet (b1) = ?

When we carpet the room,

Area of carpet = Area of room

i.e. l1*b1 = l*b

or, 12m*b1 = 7m*6m

or, b1 = 42m2/12m

or, b1 = 3.5 m

∴ The breadth of the carpet is 3.5m.

Here,

Length of stone (l1) = 40 cm = 0.4m

Breadth of stone (b1) = 30 cm = 0.3m

∴ Area of each stone (a) = l1*b1

= 0.4m * 0.3m = 0.12m2

And, length of courtyard (l) = 30m

breadth of courtayard (b) = 15m

∴ Area of courtyard (A) = l*b

= 30 m * 15 m

= 450 m2

Now,

Required number of stones (N) = Area of courtyard(A) / Area of stone(a)

= 450m2 / 0.12m2

= 3750

Hence, 3750 stones of given dimension are required to cover the courtyard.

0%

75000 m3
75 m3
7500 m3
750 m3

1200 m3
948.64 m3
1562 m3
945.25 m3

84
122
90
155

56 cm3
136 cm3
82 cm3
215 cm3

120 kg
50 kg
60 kg
250 kg

7 : 3
6 : 5
2 : 5
3 : 7

522021
6400
2400
5600

20 : 25
20 : 12
25 : 18
15 : 18

1.1 cm
1 m
1 dm
60 m

40 m3
50 m3
60 m3
44 m3

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