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