Area

Subject: Compulsory Maths

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Overview

Area is the quantity that expresses the extent of a two-dimensional figure or shape in the plane.

Area

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

Things to remember
  • There are different formulas to calculate different shapes like land, playground, water tank, circular track, etc.
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Area
Area of a Circle | MathHelp.com
Area of triangles and quadrilaterals
How To Find The Area Of A Rectangle
Maths Help Online - Area of a Trapezium
Questions and Answers

Solution:

Length of carpet(l1) = ?

Breadth of carpet (b1) = 50cm = 0.5m

Length of room (l) = 8m

Breadth of room (b) = 6m

While carpeting a room,

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

i.e. l1 * b1 = l*b

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

= \(\frac{8m*6m}{0.5m}\)

= \(\frac{48m^2}{0.5m}\)

= 96m

∴ 96m long carpet is required.

Again,

Length of carpet (l1) = 96m

Cost of carpet/metre (C) = Rs 450/m

Toal cast (T) = l1 * C

= 96 * Rs 450

Rs 43,200

Hence, it costs Rs 43,200 in order to carpet the room.

Solution:

The given window frame consists of a rectangle and a semi-circle.

For semi-circle,

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 (A1) = l * b

= 280cm * 150cm

= 42,000cm2

Area of semi circle (A2) =\(\frac{1}{2}\)\(\pi\)r2

=\(\frac{1}{2}\)*\(\frac{22}{7}\)*(140cm)2

= 30,800cm2

∴ Area of window frame (A) = A1 + A2

= 42,000cm2 + 30,800cm2

= 72,800cm2

Hence,

Perimeter of window frame (P) = 1020cm

Area ofwindow frame (A) = 72,800cm2 = 7.28m2

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:

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

= 250m2

Again,

Area of WXYZ = b * d

= 30m * 5m

= 150m2

Also,

Area of crossing path ABCD = d2

= (5m)2

= 25m2

Hence,

Area of shaded region = Area of EFGH + Area of WXYZ- Area of crossing path ABCD

= 250m2 + 150m2- 25m2

= 375m2

Alternatively,

Area of crossing paths/shaded region = d(l + b - d)

= 5m (50m - 30 - 5m)

= 5m * 15m

= 375m2

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.

Solution:

Given,

Length of wall (l) = 8m

Breadth of wall (b) = 6m

Height of wall (h) = 4m

Length of door (l1) = 2m

Height of door (h1) = 1.5m

Length of window (l2) = 2m

Height of window (h2) = 3m

Cost of painting walls (C) = Rs 80/m2

Now,

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

= 2 × 4m (8m + 6m)

= 8m × 14m

= 112m2

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

= 2 × 4m (8m + 6m) + 8m × 6m

= 8m × 14m + 48m2

= 112m2 + 48m2

= 160m2

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

= 2 × 4m (8m + 6m) -2m × 1.5m - 2(2m × 3m)

=8m × 14m - 3m2 - 2 × 6m2

= 112m2 - 3m2 - 12m2

= 97m2

Cost of painting 4 walls and ceiling = C * A

= Rs 80/m2 × 97m2

= Rs 7760

Hence,

Area of 4 walls = 112m2

Area of 4 walls and ceiling = 160m2

Area of 4 walls excluding door and windows = 97m2

Cost of painting 4 walls and ceiling = Rs 7760

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.

Here,

Length of carpet (l1) = ?

Breadth of carpet (b1) = 50cm = 0.5m

Length of room (l) = 8m

Breadth of room (b) = 6m

While carpeting a room,

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

i.e. l1*b1= l*b

or, l1 = l*b / b1

= 8m * 6m / 0.5 m

= 96m

∴ 96m long carpet is required.

Again,

Length of carpet (l1) = 96m

cost of carpet/ metre (C) = Rs.400/m

Total cost (T) = l1*C

= 96 * Rs.400

= Rs.38400

Hence, it costs Rs.38400 in order to carpet the room.

Hence,

length (l) = 40m

breadth (b) = 30 m

area of 4 walls (A) = 600m2

height (h) =?

Area of 4 walls (A) = 400m2

i.e. 2h(l + b) = 400 m2

or, 2h( 30m + 20m) = 400m2

or, 2h * 50m = 400m2

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

= 400m2 + 600m2

= 1000m2

b) Area of 4 walls, ceiling and floor (T.S.A)

= 2h (l + b) + 2lb

= 2*4m (30m + 20m ) + 2*30m*20m

= 400m2 + 1200m2

= 1600m2

c) Area of floor and ceiling

= 2 lb

= 2*30m*20m

=1200m2

d) Area of 4 walls and ceiling (A) = 1000m2

cost of painting the wall (C) = Rs.200/m2

Total cost (T) = C*A

= Rs.200/m2*1000m2

= Rs.200000

Total cost of painting four wall and ceiling is Rs.200000.

Here,

length (l) = 5m

breadth (b) = 4 m

height (h) = 3m

Now,

a) Area of floor (A)

= l*b

= 5m * 4m

= 20m2

b) Area of 4 walls (A)

= 2h (l + b)

= 2*3m (5m + 4m)

= 54m2

c) Area of ceiling (A)

= l * b

= 5m * 4m

= 20m2

d) Area of 4 wall and ceiling (A)

= 2h (l + b) + lb

= 2*3m (5m + 4m) + 5m * 4m

= 54 m2+ 20m2

= 74 m2

e) Area of 4 wall, ceiling and floor (TSA)

= 2h (l + b) + lb + lb

= 2*3m(5m + 4m) + 5m * 4m + 5m * 4m

=54m2 + 20m2+ 20m2

=94m2

Quiz

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