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Class VII - Maths: Area and Perimeter
One Word Answer Questions:
Q) Find the area of a square of side 8 cm?
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Q) 2.3 hectares in ares?
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Q) Find the area of the square?
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Q) Find the area of a rectangle of length 18 cm and breadth 9 cm?
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Q) Find the length of a rectangle whose breadth is 6 cm and area is 80 sq. cm?
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Short Answer Questions:
Q) Find the area of a rectangle whose length is 50 m and breadth is 450 cm?
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Q) Find the area of the following figure?
    
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Q) .(a)Find the cost of painting a wall of length 70 m and breadth 18 m at the rate of $ 4 per sq. m? (b)Find the number of bricks to be laid in a square path of side 12 cm, if the side of each brick is 5 cm?
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Q) The floor of building consists of 2500 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is $8?
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Q) The point A, B, C have respective co-ordinates (2, 5), (-7, 4) and (8, -6). Find the area of triangle ABC and the length of the perpendicular from A on BC?
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Long Answer Questions:
Q) Find the area of the rhombus having each side equal to 20 cm and one of its diagonals equal to 22 cm?
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Q) (a)Find the area of the rhombus having each side equal to 20 cm and one of its diagonals equal to 22 cm? (b)The point A, B, C, D have respective co-ordinates (-2, -3), (6, -5) , (18, 9) and (0, 12). Find the area of quadrilateral ABC?
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Q) (a)The polar co-ordinates of the vertices of a triangle are (-a, ?/6), (a, ?/2) and (-2a, - 2?/3) find the area of the triangle? (b)The centre of a circle is at (2, 6) and a chord of this circle of length 24 units is bisected at (- 1, 2). Find the radius of the circle?
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Q) If the co-ordinates of the vertices of a triangle ABC be (3, 0), (0, 6) and (6, 9) and if D and E divide AB and AC, respectively internally in the ratio 1 : 2, then show that the area of triangle ABC = 9. the area of triangle ADE?
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Q) The sides of the triangular plot are in the ratio 4 : 2 : 6 and the perimeter is 190 m. Find its area?
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  • Length of a rectangle = Area ÷ breadth
    l = A ÷ b
  • Breadth of a rectangle = Area ÷ length
    b = A ÷ l
  • area of the square = `( side )^2`
  • The measure of a region is called its area.
  • Conversion Table:
    1 m. = 100 cm.
    1 sq. m. = 10000 sq. cm.
    1 cm. = 10 mm.
    1 sq. cm. = 100 sq. mm.
    1 km = 1000 m
  • Length ∗ breadth = Area of the rectangle
  • Area of rectangle = length ∗ breadth = l ∗ b sq. units
  • Length of the rectangle = Area of the Rectangle x Breadth of the Rectangle
  • Breadth of the rectangle = Area of the Rectangle x Length of the Rectangle
  • Area of a square = (side ∗ side) square units.
  • Area of rhombus ABCD = 2 Area of Δ ABC
    = base x height sq. units
  • area of rhombus = 4 ∗ area of Δ AOB
    = 1/2 ∗ d1 ∗ d2; where d1 and d2 are diagonals.
  • area of rhombus = `1/2` (product of diagonals) square units
  • Perimeter of rhombus = 4 ∗ side
  • The area of a triangle formed by joining the points (x1, y2), (x2, y2) and (x2, y3) is `1/2` |y1 (x2 - x3) + y2 (x3 - x1) + y3 (x1 - x2)| sq. units
  • If a, b, c are the sides of the triangle, then the perimeter of triangle = (a + b + c) units.
  • Area of the triangle = `(sqrt(s(s - a) (s - h) (s - c)) )`
  • The semi-perimeter of the triangle, s =` ((a + b + c)/2)`
  • Area of triangle = `(1/2)` ∗ base ∗ height
  • Base of the triangle = `((2 Area)/[height])`
  • Height of the triangle = `((2 Area)/[base])`
  • If a represents the side of an equilateral triangle, then its area = `((a^2)(sqrt3)/4)`
  • Area of right angled triangle
    = `(1/2)` ∗ b ∗ h
  • Area of the triangle = `((sqrt(3)/4) a^2 )`square units
  • Height of the triangle = `((sqrt3)/2)` a units
  • Area of the triangle = `(1/2)` ∗ base ∗ height
  • Circumference/Diameter = Pie
  • c/d = π or c = πd
  • Diameter is twice the radius, i.e., d = 2r
    C = 2πr
  • value of π = 22/7 or 3.14
  • Circumference of circle = 2πr
  • Area of circle = `πr^2`
  • Squares And Rectangles
  • Rectangles
  • squares
  • Area
  • Interior and Exterior of a Region
  • Measuring Regions
  • Area of a Rectangle
  • Squares
  • Solved examples to find the area of a square when side is given
  • To Find Area of a Rectangle when Length and Breadth are of Different Units
  • Areas of Irregular Figures
  • Area of Rhombus
  • Worked-out examples on area of rhombus
  • Area of Triangle
  • Worked-out problems to find the area of a triangle given 3 points
  • Alternative Method
  • Area and Perimeter of the Triangle
  • Area of right angled triangle
  • Worked-out examples on area and perimeter of the triangle
  • More solved examples on area and perimeter of the triangle
  • Circle
  • Area of circle
  • Worked-out examples on how do you find the area of a circle and the circumference of circle
  • Units of Area conversions
  • Area Conversion
  • Relation between the various units of area
  • Examples on area conversion
Squares And Rectangles

Rectangles

A 4-sided flat shape with straight sides where all interior angles are right angles (90°). Also opposite sides are parallel and of equal length.
Example: A square is a special type of rectangle.

periandarea

To find Length or Breadth when Area of a Rectangle

When we need to find length of a rectangle we need to divide area by breadth.
Length of a rectangle = Area ÷ breadth
l = A ÷ b

Similarly, when we need to find breadth of a rectangle we need to divide area by length.
Breadth of a rectangle = Area ÷ length
b = A ÷ l

Formulas:
l = A ÷ b
b = A ÷ l


For Example:

1. Find the length of a rectangle whose breadth is 5 cm and area is 90 sq. cm.

Solution:

  • Breadth (b) = 5 cm,
    Area (A) = 90 sq. cm.
    Length (l) = Area ÷ breadth
    = 90 ÷ 5
    = 18
  • Therefore, length of rectangle = 18 cm.


2. Find the breadth of a rectangle whose length is 200 cm and area is 10 `m^2`.

Solution:

  • Length (l) = 200 cm,
    Area (A) = 10 `m^2`
    100 cm = 1 m.
    Length (l) = Area ÷ breadth
    Length (l) = 200/100 m = 2 m.
    Breadth (b) = A ÷ l
    = 10 ÷ 2 m.
    = 5 m.

Therefore, breadth of rectangle = 5 m.


squares

A 4-sided flat shape with straight sides where:

  • All sides have equal length, and
  • Every interior angle is a right angle (90°)
It is a Quadrilateral and a Regular Polygon

periandarea

Problem 1) .
Find the area of the square?

Solution:

periandarea

consider a square of side of 'a' units and its area 25 square units.
area of the square = ( side )2
25 = a2
(or)
a2 = 25
a= +5 0r -5

Area

The amount of surface that a plane figure covers is called its area. It’s unit is square centimetres or square meters etc.
A rectangle, a square, a triangle and a circle are all examples of closed plane figures.

Interior and Exterior of a Region

The part of the plane enclosed by a closed figure is called the interior region and the part of the plane outside the enclosed figure is called the exterior region.

periandarea

In the adjacent figure a rectangle ABCD is shown. It’s interior and exterior regions are also shown here.
Area is always measured in squares and the unit of area square units (sq. cm., sq. m. or `cm^2`, `m^2`).

Measuring Regions:

We measure a given region by a unit region and find how many such unit regions are contained in the given region.

periandarea

The measure of a region is called its area.
The area is always expressed in square units. The area of a square with side 1 cm by 1 cm is equal to 1 square cm.

periandarea periandarea

Area is a measure of any area surface, e.g., the surface of a table, the surface of your pencil box etc.
Area is two dimensional.

It means, to find the area of any surface we need to know two sides.
Note: Here we’ll discuss only areas of squares and rectangles. Given below is a table of units of sides and corresponding units for areas.

 Side   Area 
 cm   sq.mm or mm2 
 cm   sq.cm or cm2 
 m   m or m2 
 km   sq.km or km2 

Conversion Table:

  • 1 m. = 100 cm.
    1 sq. m. = 10000 sq. cm.
    1 cm. = 10 mm.
    1 sq. cm. = 100 sq. mm.
    1 km = 1000 m

For finding the area of a given figure, make sure that the sides (length or breadth) are in the same unit of length. If they are given in different units, change them to the same unit.
Area of a Rectangle:

Area of a rectangle is discussed here. We know, that a rectangle has length and breadth.
Let us look at the rectangle given below.
Each rectangle is made of squares. The side of each square is 1 cm long. The area of each square is 1 square centimetre. The rectangle ABCD has 8 such squares. Therefore, its area is 8 sq cm. Similarly we can find the areas of the other rectangles by counting the number of squares. We also note the length and breadth of each rectangle and write in the table below:

Rectangle Area Length Breadth Length X Breadth
ABCD 8 Sq.cm 4 cm 2 cm 4 cm x 2 cm = 8 cm2
LMNO 12 Sq.cm 4 cm 3 cm 4 cm x 3 cm = 12 cm2
PQRS 6 Sq.cm 2 cm 3 cm 2 cm x 3 cm = 6 cm2

In each case we observe the length ∗ breadth = Area of the rectangle.
Therefore, area of rectangle = length ∗ breadth = l ∗ b sq. units
From the above multiplication, we get the following facts:

  • Length of the rectangle = Area of the Rectangle x Breadth of the Rectangle
    Breadth of the rectangle = Area of the Rectangle x Length of the Rectangle

Solved examples to find the area of a rectangle when length and breadth are given:
1. Find the area of a rectangle of length 12 cm and breadth 3 cm.

  • Length (l) of the rectangle = 12 cm.
    Breadth (b) of the rectangle = 3 cm.
    Area of the rectangle = length ∗ breadth
    = 12 ∗ 3 sq cm.
    = 36 sq cm.

2. Find the area of a rectangle of length 15 cm and breadth 6 cm.

  • Length of the rectangle = 15 cm.
    Breadth of the rectangle = 6 cm.
    Area of the rectangle = l ∗ b
    = 15 ∗ 6 sq cm.
    = 90 sq cm.

3. Find the area of a rectangle of length 17 cm and breadth 9 cm.

  • Length of the rectangle = 17 cm.
    Breadth of the rectangle = 9 cm.
    Area of the rectangle = l ∗ b
    = 17 ∗ 9 sq cm.
    = 153 sq cm.

4. Find the area of a rectangle of length 24 mm and breadth 8 mm.

  • Length of the rectangle = 24 mm.
    Breadth of the rectangle = 8 mm.
    Area of the rectangle = l ∗ b
    = 24 ∗ 8 sq mm.
    = 192 sq mm.

5. Find the area of a rectangle of length 37 mm and breadth 19 mm.

  • Length of the rectangle = 37 mm.
    Breadth of the rectangle = 19 mm.
    Area of the rectangle = l ∗ b
    = 37 ∗ 19 sq mm.
    = 703 sq mm.
Squares
  • In area of a square we will learn how to find the area by counting squares.
  • To find the area of a region of a closed plane figure, we draw the figure on a centimetre squared paper and then count the number of squares enclosed by the figure.
  • We know, that square is a rectangle whose length and breadth are equal. in a square all four sides are equal.
  • Therefore, area of a square = (side ∗ side) square units.
Solved examples to find the area of a square when side is given:

1. Find the area of a square of side 8 cm.

  • Area of a square = length (l) ∗ length (l)
    = 8 ∗ 8 sq. cm.
    = 64 sq. cm.

2. Find the area of a square of side 11 m.

  • Area of a square = length ∗ length
    = 11 ∗ 11 sq. m.
    = 121 sq. m.

3. Find the area of a square of side 49 cm.

  • Area of a square = length ∗ length
    = 49 ∗ 49 sq. cm.
    = 2401 sq. cm.

4. Find the area of a square of side 50 m.

  • Area of a square = length ∗ length
    = 50 ∗ 50 sq. m.
    = 2500 sq. m.
To Find Area of a Rectangle when Length and Breadth are of Different Units:

For Example:

1. Find the area of a rectangle whose length is 15 cm and breadth is 50 mm.

Solution:

  • Length (l) = 15 cm.
    10 mm = 1 cm.
    Therefore, 50 mm = 50/10 cm = 5 cm.
    Breadth = 5 cm.

  • Area = l ∗ b
    = 15 ∗ 5 sq. cm.
    = 75 sq. cm.

2. Find the area of a rectangle whose length is 32 m and breadth is 200 cm.

Solution:

  • Length (l) = 32 m.
    100 cm = 1 m.
    Therefore, 200 cm = 200/100 m = 2 m.
    Breadth = 2 m.

  • Area = l ∗ b
    = 32 ∗ 2 sq. m.
    = 64 sq. m.

3. Find the area of a rectangle whose length is 17 cm and breadth is 85 mm.

Solution:

  • Length (l) = 17 cm.
    10 mm = 1 cm.
    Therefore, 85 mm = 85/10 cm= 8.5 cm.
    Breadth = 8.5 cm.

  • Area = l ∗ b
    = 17 ∗ 8.5 sq. cm.
    = 144.5 sq. cm.

4. Find the area of a rectangle whose length is 21 m and breadth is 500 cm.

Solution:

  • Length (l) = 21 m.
    100 cm = 1 m.
    Therefore, 500 cm = 500/100 m = 5 m.
    Breadth = 5 m.

  • Area = l ∗ b
    = 21 ∗ 5 sq. m.
    = 105 sq. m.

5. Find the area of a rectangle whose length is 41 m and breadth is 400 cm.

Solution:

  • Length (l) = 41 m.
    100 cm = 1 m.
    Therefore, 400 cm = 400/100 m = 4 m.
    Breadth = 4 m.

  • Area = l ∗ b
    = 41 ∗ 4 sq. m.
    = 164 sq. m.

To find Length or Breadth when Area of a Rectangle is given:

When we need to find length of a rectangle we need to divide area by breadth.
Length of a rectangle = Area ÷ breadth
l = A ÷ b
Similarly, when we need to find breadth of a rectangle we need to divide area by length.
Breadth of a rectangle = Area ÷ length
b = A ÷ l

Formulas:
For Example:

1. Find the length of a rectangle whose breadth is 5 cm and area is 90 sq. cm.

Solution:

  • Breadth (b) = 5 cm,
    Area (A) = 90 sq. cm.
    Length (l) = Area ÷ breadth
    = 90 ÷ 5
    = 18
  • Therefore, length of rectangle = 18 cm.

2. Find the breadth of a rectangle whose length is 200 cm and area is 10 m^2.

Solution:

  • Length (l) = 200 cm,
    Area (A) = 10 `m^2`
    100 cm = 1 m.
    Length (l) = Area ÷ breadth
    Length (l) = 200/100 m = 2 m.
    Breadth (b) = A ÷ l
    = 10 ÷ 2 m.
    = 5 m.
  • Therefore, breadth of rectangle = 5 m.

Areas of Irregular Figures

periandarea

Solution:

Area of a rectangle ABDC = 3 ∗ 1
= 3 sq. cm.

Area of a rectangle EFGD = 2 ∗ 1
= 2 sq. cm.

Therefore, Total Area = 3 + 2
= 5 sq. cm.

Area of the given figure = 5 sq. cm.

2. Find the area of the following figure.

periandarea
  • Total area = Area of the rectangle ABGF + Area of the rectangle CDEG
    = 8 ∗ 2 cm2+ 2 ∗ (8 - 2) cm2
    = 16 sq cm2+ 2 ∗ 6 cm2
    = (16 + 12) cm2
    = 28 cm2

Therefore, area of the figure = 28 cm2

3. Find the area of the figure given on the right side.

periandarea
  • Total area = Area of the rectangle ABKL + Area of the rectangle EFGH + Area of the rectangle CDIJ
    = 20 ∗ 4 cm2 + 20 ∗ 4 cm2 + 8 ∗ 4 cm2
    = 80 cm2 + 80 cm2 + 32 cm2
    = (80 + 80 + 32) cm2
    = 192 cm2
  • Therefore, area of the figure = 192 cm2


To find Cost of Painting or Tilling when Area and Cost per Unit is given:

To find the cost, find the area and multiply it by cost per sq. unit.

Cost of tiling or painting = area ∗ cost per square unit


For Example:

1. Find the cost of painting a wall of length 80 m and breadth 15 m at the rate of $ 4 per sq. m.

Solution:

    Area of wall = length ∗ breadth

  • Here, length = 80 m and breadth = 15 m.
    = 80 ∗ 15 sq. m.
    = 1200 sq. m.
  • Therefore, painting the wall = area ∗ cost per square unit.

  • Given, rate = $ 4 per sq. m.
    = 1200 ∗ 4
    = $ 4800.

2. Find the cost of painting a house of length 160 m and breadth 120 m at the rate of $ 5 per sq. m.

Solution:

    Area of wall = length ∗ breadth

  • Here, length = 160 m and breadth = 120 m.
    = 160 ∗ 120 sq. m.
    = 19200 sq. m.
  • Therefore, painting the wall = area ∗ cost per square unit.

  • Given, rate = $ 5 per sq. m.
    = `19200 / 5`
    = $ 3840.

To find the Number of Bricks or Tiles when Area of Path and Brick is given:

Number of bricks or tiles = area of path or wall ÷ area of a brick or tile.

For Example:

1. Find the number of bricks to be laid in a square path of side 18 cm, if the side of each brick is 3 cm.

Solution:

Area of path = 18 ∗ 18 sq. cm.
Area of each brick = 3 ∗ 3 sq. cm.

  • Therefore, number of bricks = area of path ÷ area of each brick.
    = (18 ∗ 18) ÷ (3 ∗ 3)
    = 324 ÷ 9
    = 36

Therefore, number of bricks = 36.


2. Find the number of bricks to be laid in a square path of side 25 cm, if the side of each brick is 5 cm.

Solution:

    Area of path = 25 ∗ 25 sq. cm.

    Area of each brick = 5 ∗ 5 sq. cm.

  • Therefore, number of bricks = area of path ÷ area of each brick.
    = (25 ∗ 25) ÷ (5 ∗ 5)
    = 625 ÷ 25
    = 25
  • Therefore, number of bricks = 25.

Area of Rhombus

periandarea

ABCD is a rhombus whose base AB = b, DB ⊥ AC DB = d1 AC = d2 and the altitude from C on AB is CE, i.e., h.

Area of rhombus ABCD = 2 Area of Δ ABC
                                        = 2 ∗ `1/2` AB ∗ CD sq units.
                                        = 2 ∗ `1/2` ∗ b ∗ h sq. units
                                        = base x height sq. units

Also, area of rhombus = 4 ∗ area of Δ AOB
                                        = 4 ∗ `1/2` ∗ AO ∗ OB sq. units
                                        = 4 ∗ `1/2` ∗ `1/2`∗ d2 ∗ `1/2`∗ d1 sq. units
                                        = 4 ∗` 1/8`∗ d1 ∗ d2 square units
                                        = `1/2` ∗ d1 ∗ d2; where d1 and d2 are diagonals.

Therefore, area of rhombus = `1/2` (product of diagonals) square units
Perimeter of rhombus = 4 ∗ side


Worked-out examples on area of rhombus:

1. Find the area of the rhombus having each side equal to 17 cm and one of its diagonals equal to 16 cm.

Solution:

    periandarea
  • ABCD is a rhombus in which AB = BC = CD = DA = 17 cm
    AC = 16 cm
    Therefore, AO = 8 cm
    In Δ AOD,
    AD2= AO2+ OD2
    ⇒ 172 = 82 + OD2
    ⇒ 289 = 64 + OD2
    ⇒ 225 = OD2
    ⇒ OD = 15

  • Therefore, BD = 2 OD
    = 2 ∗ 15
    = 30 cm

  • Now, area of rhombus
    = `1/2` ∗ d1 ∗ d2
    = `1/2` ∗ 16 ∗ 30
    = 240 cm2

2. Find the altitude of the rhombus whose area is 315 cm2 and its perimeter is 180 cm.

Solution:

  • Since, the perimeter of rhombus = 180 cm
    Therefore, side of rhombus = P/4 = 180/4 = 45 cm
    Now, area of rhombus = b ∗ h
    ⇒ 315 = 45 ∗ h
    ⇒ h = 315/45
    ⇒ h =7 cm
  • Therefore, altitude of the rhombus is 7 cm.


3. The floor of building consists of 2000 tiles which are rhombus shaped and each of its diagonals are 40 cm and 25 cm in length. Find the total cost of polishing the floor, if the cost per m2 is $5.

Solution:

  • In each rhombus, the tile length of the diagonals = 40 cm and 25 cm
    Therefore, area of each tile = `1/2` ∗ 40 ∗ 25 = 500 cm2
    Therefore, area of 2000 tiles = 2000 ∗ 500 cm2
    = 1000000 cm2
    = 1000000/10000 cm2
    = 100 m2
  • For 1 m2cost of polishing = $5
    = $5 ∗ 100 = $500.

The formula of perimeter and area of rhombus are explained above with the detailed step-by-step explanation using various examples.


Area of Triangle

The area of a triangle formed by joining the points (x1, y2), (x2, y2) and (x2, y3) is
`1/2`|y1 (x2 - x3) + y2 (x3 - x1) + y3 (x1 - x2)| sq. units



Worked-out problems to find the area of a triangle given 3 points


1. Find the value of x for which the area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is `12 1/2` sq. units.

Solution:

    The area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is
    `1/2`|(- 1 - 4x - 4x) - (- 4x + x + 4)|
    = `1/2` |- 1 - 8x + 3x - 41 = 1/2 |- 5x - 5| sq. units.
    By problem, `1/2` |-1 - 5x - 5| = `121/2 = 25/2`

  • Therefore, 5x + 5 = ± 25
    or, x + 1 = ± 5
  • Therefore, x = 4 or, - 6.


2. The point A, B, C have respective co-ordinates (3, 4), (-4, 3) and (8, -6). Find the area of Δ ABC and the length of the perpendicular from A on BC.

Solution:

    The required area of the triangle ABC.

  • = `1/2`|(9 + 24 + 32) - (- 16 + 24 - 18)| sq. unites.
    = `1/2` |65 + 10| sq. units = `75/2` sq. units.
  • Again, BC = distance between the points B and C
    = `sqrt(([(8 + 4)^2)+( (- 6 - 3)^2)])` = `sqrt(144 + 81)` = `sqrt(225)` = 15 units.
    Let p be the required length of the perpendicular from A on BC then,

  • `1/2` • BC • p = area of the triangle ABC
    or, `1/2` • 15 • p = `75/2`
    or, p = 5
  • Therefore, the required length of the perpendicular from A on BC is 5 units.


3. The point A, B, C, D have respective co-ordinates (-2, -3), (6, -5) , (18, 9) and (0, 12). Find the area of quadrilateral ABC.

Solution:

  • We have, the area of the triangle ABC
    = `1/2` |(10 + 54 - 54) - (- 18 - 90 - 18)| sq. units
    = `1/2` (10 + 126) sq. units
    = 68 sq. units.

  • Again, area of the triangle ACD
  • = `1/2` |(- 18 + 216 + 0) - (- 54 + 0 - 24)|sq. units
    = `1/2` (198 + 78) sq. units
    = 138 sq. units.
  • Therefore, the required area of the quadrilateral ABCD
    = area of the Δ ABC + area of the ΔACD
    = (68 + 138) sq. units
    = 206 sq. units.
Alternative Method

[This method is analogous with the short-cut method of getting the area of a triangle. Suppose, we want to find the area of the quadrilateral whose vertices have co-ordinates (x1, y1), (x2, y2), (x3, y3) and (x4, y4). For this, we write the co-ordinates of the vertices in four rows repeating the first written co-ordinates in the fifth row. Now take the sum of the products of digits shown by (↘) and from this sum subtract the sum of the products of digits shown by (↗). The required area of the quadrilateral will be equal to half of the difference obtained. Thus, the area of the quadrilateral
`1/2` |(x1y2 + x2 y3 + x3y4 + x4y1) - (x2y1 + x3y2 + x4y3 + x1y4)| sq. units.
The above method can be used to find the area of a polygon of any number of sides when the co-ordinates of its vertices are given.]

Solution:

    The required area of the quadrilateral ABCD

  • = `1/2` |(10 + 54 + 216 + 0) - (- 18 - 90 + 0 - 24)| sq. units.
    = `1/2` (280 + 132) sq. units.
    = `1/2` ∗ 412 sq. units.
    = 206 sq. units.

4. The co-ordinates of the points A, B, C, D are (0, -1), (-1, 2), (15, 2) and (4, -5) respectively. Find the ratio in which AC divides BD.

Solution:

    Let us assume that the line-segment AC divides the line -segment BD in the ratio m : n at P.

    Therefore, P divides the line-segment BD in the ratio m : n. Hence, the co-ordinates of P are.
    `[(m xx 4 + n xx (-1))/(m + n), (m xx (-5) + n xx 2)/(m + n)] + [(4m - n)/(m + n), (5m + 2n)/(m + n)]`.
    Clearly, the points A, C and P are collinear. Therefore, the area of the triangle formed by the point A, C and P must be zero.

  • Therefore,`1/2[( 0 + 15 xx (- 5m + 2n))/(m + n) - (4m - n)/(m + n) - (- 15 + 2 xx (4m - n))/(m + n) + 0] = 0`
    or, `(15 xx (-5m + 2n))/(m + n) - (4m - n)/(m + n) + (15 - 2 xx (4m - n))/(m + n)=0`
    or,` - 75m + 30n - 4m + n + 15m + 15n - 8m + 2n = 0`.
    or,` - 72m + 48n = 0 `
    or, `72m = 48n`
    or,` m/n = 2/3`.
  • Therefore, the line-segment AC divides the line-segment BD internally in the ratio 2 : 3.

5. The polar co-ordinates of the vertices of a triangle are `(-a, pi/6), (a, pi/2)` and `(-2a, - (2pi)/3)` find the area of the triangle.

Solution:

    The area of the triangle formed by joining the given points

  • = `1/2 |a (-2a) sin((-2pi)/3 - pi/2) + (-2a) (-a) sin(pi/6 + (2pi)/3) - (-a) a sin(pi/6 + pi/2)| `sq. units. [ using above formula]
    = `1/2` `|2a^2 sin(π + π/6 ) + 2a^2 sin⁡ (π - π/6) -2a^2 sin⁡ (π/2 - π/6)|`sq. units
    . = `1/2` `|-2a^2 sin⁡(π/6) + 2a^2 sin⁡(π/6) - a^2 cos⁡(π/6)|` sq. units.
    = `1/2 a^2(sqrt3/2)` sq. units = `sqrt3/4) a^2`sq. units.

6. The centre of a circle is at (2, 6) and a chord of this circle of length 24 units is bisected at (- 1, 2). Find the radius of the circle.

Solution:

'

    Let C (2, 6) be the centre of the circle and its chord AB of length 24 units is bisected at D (- 1, 2).

  • Therefore, CD2= (2 + 1) 2 + (6 - 2) 2
    = 9 + 16 = 25 and DB = `1/2` • AB = `1/2` • 24 = 12
  • Join CB. Now, D is the mid-point of the chord AB; hence, CD is perpendicular to AB. Therefore, from the triangle BCD we get,

  • BC2= CD2+ BD2= 25 + 122= 25 + 144 = 169
    or, BC = 13
  • Therefore, the required radius of the circle = 13 units.

7. If the co-ordinates of the vertices of a Δ ABC be (3, 0), (0, 6) and (6, 9) and if D and E divide AB and AC, respectively internally in the ratio 1 : 2, then show that the area of Δ ABC = 9 • the area of Δ ADE.

Solution:

    By question D divides AB internally in the ratio 1 : 2; hence,
    the co-ordinates of D are `((1 ** 0 + 2 ** 3)/(1 + 2), (1 ** 6 + 2 ** 0)/(1 + 2))`
    `= (6/3, 6/3) = (2, 2)`.
    Again, E divides AC internally in the ratio 1 : 2; hence, the co-ordinates of E are
    `((1 ** 6 + 2 ** 3)/(1 + 2), (1 ** 9 + 2 ** 0)/(1 + 2))`
    `= (12/3, 9/3) = (4, 3)`.

    Now, the area of the triangle ABC

  • = `1/2` |(18 + 0 + 0) - (0 + 36 + 27)| sq. units.
    = `1/2` |18 - 63| sq. units.
    = `45/2 `sq. units.
  • And the area of the triangle ADE
    = `1/2` |( 6 + 6 + 0) - (0 + 8 + 9)| sq. units.
    = `1/2` = `5/2` sq. units.
  • therefore, area of the Δ ABC

  • = `45/2` sq. units = `9 * 5/2` sq. units.
    = 9 • area of the Δ ADE. Proved.

Area and Perimeter of the Triangle

  • If a, b, c are the sides of the triangle, then the perimeter of triangle = (a + b + c) units.
  • Area of the triangle =`sqrt(s(s - a) (s - b) (s - c))`
    The semi-perimeter of the triangle, s = `((a + b + c)/2)`
  • In a triangle if 'b' is the base and h is the height of the triangle then
    Area of triangle = `1/2 `∗ base ∗ height
  • Similarly,

    periandarea
  • Base of the triangle = `((2 Area)/[height] )`
  • Height of the triangle = `((2 Area)/[base])`
Area of right angled triangle

If a represents the side of an equilateral triangle, then
its area = `((a^2√3)/4)`

periandarea

Area of right angled triangle
A = `1/2` ∗ BC ∗ AB
= `1/2` ∗ b ∗ h

periandarea

Worked-out examples on area and perimeter of the triangle:

1. Find the area and height of an equilateral triangle of side 12 cm. (√3 = 1.73).

Solution:

  • Area of the triangle = `sqrt(3)/4` a2 square units
    = `sqrt(3)/4` ∗ 12 ∗ 12
    = 36`sqrt(3)` cm2
    = 36 ∗ 1.732 cm2
    = 62.28 cm2
  • Height of the triangle = `sqrt(3)/2 ` units
    = `sqrt(3)/2 xx 12 `cm
    = 1.73 ∗ 6 cm
    = 10.38 cm

2. Find the area of right angled triangle whose hypotenuse is 15 cm and one of the sides is 12 cm.

Solution:

  • AB2 = AC2 - BC2
    = 152 - 122
    = 225 - 144
    = 81
  • Therefore, AB = 9

  • Therefore, area of the triangle = `1/2` ∗ base ∗ height
    = `1/2` ∗ 12 ∗ 9
    = 54 cm2

3. The base and height of the triangle are in the ratio 3 : 2. If the area of the triangle is 243 cm2 find the base and height of the triangle.

Solution:

  • Let the common ratio be x
    Then height of triangle = 2x
    And the base of triangle = 3x
    Area of triangle = 243 cm2
  • Area of triangle = `1/2` ∗ b ∗ h
    243 = `1/2` ∗ 3x ∗ 2x
    ⇒ 3x2 = 243
    ⇒ x2 = `243/3`
    ⇒ x = `sqrt[81]`
    ⇒ x = `sqrt(9 xx 9)`
    ⇒ x = 9
  • Therefore, height of triangle = 2 ∗ 9
    = 18 cm

  • Base of triangle = 3x
    = 3 ∗ 9
    = 27 cm

4. Find the area of a triangle whose sides are 41 cm, 28 cm, 15 cm. Also, find the length of the altitude corresponding to the largest side of the triangle.

Solution:

  • Semi-perimeter of the triangle = `((a + b + c)/2)`
    = `((41 + 28 + 15)/2)`
    = `84/2`
    = 42 cm

  • Therefore, area of the triangle = `sqrt(s(s - a) (s - b) (s - c))`
    = `sqrt(42 (42 - 41) (42 - 28) (42 - 15)) `cm2
    = `sqrt(42 xx 1 xx 27 xx 14)` cm2
    = `sqrt(3 xx 3 xx 3 xx 3 xx 2 xx 2 xx 7 xx 7)` cm2
    = `3 xx 3 xx 2 xx 7 `cm2
    = 126 cm2

  • Now, area of triangle = `1/2` ∗ b ∗ h

  • Therefore, h = `[2A]/41`
    = `(2 xx 126)/41`
    = `252/41`
    = 6.1 cm

More solved examples on area and perimeter of the triangle:

5. Find the area of a triangle, two sides of which are 40 cm and 24 cm and the perimeter is 96 cm.
Solution:

    Since, the perimeter = 96 cm
    a = 40 cm, b = 24 cm

  • Therefore, C = P - (a + b)
    = 96 - (40 + 24)
    = 96 - 64
    = 32 cm
  • Therefore, `S = (a + b + c)/2`
    = `(32 + 24 + 40)/2`
    = `96/2`
    = 48 cm
  • Therefore, area of triangle = `sqrt(s(s - a) (s - b) (s - c))`
    =`sqrt(48 (48 - 40) (48 - 24) (48 - 32)) `
    = `sqrt(48 ∗ 8 ∗ 24 ∗ 16 )`
    = `sqrt(2 ∗ 2 ∗ 2 ∗ 2 ∗ 3 ∗ 2 ∗ 2 ∗ 2 ∗ 2 ∗ 2 ∗ 2 ∗ 3 ∗ 2 ∗ 2 ∗ 2 ∗ 2)`
    = 3 ∗ 2 ∗ 2 ∗ 2 ∗ 2 ∗ 2 ∗ 2 ∗ 2
    = 384 cm2

6. The sides of the triangular plot are in the ratio 2 : 3 : 4 and the perimeter is 180 m. Find its area.
Solution:

  • Let the common ratio be x,
    then the three sides of triangle are 2x, 3x, 4x
    Now, perimeter = 180 m
  • Therefore, 2x + 3x + 4x = 180
    ⇒ 9x = 180
    ⇒ x = 180/9
    ⇒ x = 20

  • Therefore, 2x = 2 ∗ 20 = 40
    3x = 3 ∗ 20 = 60
    4x = 4 ∗ 20 = 80

  • Area of triangle = `sqrt(s(s - a) (s - b) (s - c))`
    = `sqrt(90(90 - 80) (90 - 60) (90 - 40))`
    = `sqrt(90 ∗ 10 ∗ 30 ∗ 50))`
    = `sqrt(3 ∗ 3 ∗ 2 ∗ 5 ∗ 2 ∗ 5 ∗ 3 ∗ 2 ∗ 5 ∗ 5 ∗ 5 ∗ 2)`
    = 3 ∗ 2 ∗ 5 ∗ 2 ∗ 5 `sqrt(3 ∗ 5)`
    = 300 `sqrt(15) m^2`
    = 300 ∗ 3.872 m2
    = 1161.600 m2
    = 1161.6 m2
Circle
Area of circle

Circumference of circle: The distance around the circular region is called its circumference. The ratio of circumference of any circle to its diameter is constant. This constant is denoted by Π and is read as pie.

Circumference/Diameter = Pie
i.e., c/d = Π or c = Πd
We know that diameter is twice the radius, i.e., d = 2r
C = Π ∗ 2r
⇒ C = 2Πr
Therefore approximate value of Π = `22/7` or 3.14.

Area of circle: The measure of the region enclosed inside the circle is called its area.

periandarea

In case of concentric circles: The region enclosed between two concentric circles of different radii is called the area of the ring.

periandarea

Note:

Circles having same centre but different radii are called concentric circles.

Worked-out examples on how do you find the area of a circle and the circumference of circle:

1. Find the circumference and area of radius 7 cm.

Solution:

  • Circumference of circle = 2Πr
    = 2 ∗ `22/7` ∗ 7
    = 44 cm

  • Area of circle = Πr2
    = 22/7 ∗ 7 ∗ 7 cm2
    = 154 cm2

2. A race track is in the form of a ring whose inner circumference is 220 m and outer circumference is 308 m. Find the width of the track.

Solution:

  • Let r1 and r2 be the outer and inner radii of ring.
    Then 2Πr1 = 308
    2 ∗ `22/7` r1 = 308
    ⇒ r1 = `((308 ∗ 7)/(2 ∗ 22))`
    ⇒ r1 = 49 m
    2Πr2 = 220
    2 ∗ 22/7 ∗ r2 = 220
    r2 = `(220 ∗ 7)/(2 ∗ 22)`
    r2 = 35 m
  • Therefore, width of the track = (49 - 35) m = 14 m

3. The area of a circle is 616 cm2. Find its circumference.

Solution:

  • We know area of circle = Πr2
    ⇒ 22/7 ∗ r2= 616
    ⇒ r2 = (616 ∗ 7)/22
    ⇒ r2 = 28 ∗ 7
    ⇒ r = `sqrt(28 ∗ 7)`
    ⇒ r = `sqrt(2 ∗ 2 ∗ 7 ∗ 7)`
    ⇒ r = 2 ∗ 7
    ⇒ r = 14 cm

  • Therefore, circumference of circle = 2Πr
    = `2 ∗ 22/7 ∗ 14 ` = 88 cm

4. Find the area of the circle if its circumference is 132 cm.

Solution:

    We know that the circumference of circle = 2Πr

  • Area of circle = Πr2
    Circumference = 2Πr = 132
    ⇒ 2 ∗ 22/7 ∗ r = 132
    ⇒ r = (7 ∗ 132)/(2 ∗ 22)
    ⇒ r = 21 cm

  • Therefore, area of circle = Πr2
    = 22/7 ∗ 21 ∗ 21
    = 1386 cm2

5. The ratio of areas of two wheels is 25 : 49. Find the ratio of their radii.

Solution:

  • If A1 and A2 are the area of wheels,
    A1/A2 =25/49
    (Πr12)/(Πr22) = 25/49
    (r12)/(r22) = 25/49
    r1/r2 = `sqrt(25/49)`
    r1/r2 = 5/7

  • Therefore, ratio of their radii is 5 : 7.

6. The diameter of a wheel of a motorcycle is 63 cm. How many revolutions will it make to travel 99 km?

Solution:
The diameter of the wheel of a motorcycle = 63 cm
Therefore, circumference of the wheel of motorcycle = Πd
= 22/7 ∗ 63 = 198 cm

Total distance travelled by motorcycle = 99 km
= 99 ∗ 1000
= 99 ∗ 1000 ∗ 100 cm

Therefore, number of revolutions = (99 ∗ 1000 ∗ 100)/198 = 50000

7. The diameter of a wheel of cycle is 21 cm. It moves slowly along a road. How far will it go in 500 revolutions?

Solution:
In revolution, distance that wheel covers = circumference of wheel Diameter of wheel = 21 cm
Therefore, circumference of wheel = Πd
= 22/7 ∗ 21
= 66 cm
So, in 1 revolution distance covered = 66 cm
In 500 revolution distance covered = 66 ∗ 500 cm
= 33000 cm
= 33000/100 m
= 330 m

8. The circumference of a circle exceeds the diameter by 20 cm. Find the radius of the circle.
Solution:

  • Let the radius of circle of = r m.
    Then circumference = 2 Πr
    Since, circumference exceeds diameter by 20
    Therefore, according to question;
    2 Πr = d + 20
    2 Πr = 2r + 20
    2 ∗ (22/7) ∗ r = 2r + 20
    44r/7 - 2r = 20
    (44r - 14r)/7 = 20
    30r/7 = 20
    ⇒ r = (7 ∗ 20)/30
    ⇒ r =14/3
  • So, the radius of circle = 14/3 cm = 42/3 cm

9. A piece of wire in the form of rectangle 40 cm long and 26 cm wide is again bent to form a circle. Find the radius of the circle.

Solution:

Length of wire = Perimeter of rectangle
= 2(l + b)
= 2(40 + 26)
= 2 ∗ 66
= 132 cm

When it is again bent to form a circle, then
Perimeter of circle = Perimeter of rectangle
2 Πr = 132 cm
2 ∗ 22/7 ∗ r = 132
r = (132 ∗ 7)/(2 ∗ 22)
r = 21 cm

Units of Area conversions
Area Conversion:

In units of area conversion we will learn how to convert the units. It’s very important to know the relation between the various units while solving the questions on area conversion.

Relation between the various units of area

Length Units Area Units
1m= 100cm so, 1 m2=(100 x 100)cm2=10000cm2
1m= 10dm so, 1 m2=(10 x 10)dm2=100dm2
1 cm= 10mm so, 1 cm2=(10 x 10)mm2=100m2
1 km= 1000m so, 1 km2=(1000 x 1000)m2=1000000m2
1 hm= 100m so, 1 hm2=(100 x 100)m2=10000m2
( 1 hectare )
1 dam= 10m so, 1 dam2=(10 x 10)m2=100m2
( 1 ares )

Notes:

1 hectare = 100 ares

Examples on area conversion

Convert the following units:

(a) 60 cm2 in mm2

Solution:

  • 60 cm2 in mm2
    1 cm = 10 mm
    1 cm2 = 100 mm2
    Therefore, 60 cm2 = 60 ∗ 100 mm2
    = 6000 mm2

(b) 3 hectares in m2

Solution:

  • 3 hectares in m2
    1 hectare = 10000 m2
    Therefore, 3 hectares = 3 ∗ 10000 m2
    = 30000 m2

(c) 12000 m2 in hectares

Solution:

  • 12000 m2 in hectares
    10000 m2 = 1 hectare
    Therefore, 12000 m2 = 1/10000 ∗ 12000 hectare
    = 12/10 hectare
    = 6/5 hectare

(d) 2.3 hectares in areas

Solution:

  • 2.3 hectares in area
    1 hectare = 100 area
    Therefore, 2.3 hectares = 2.3 ∗ 100 area
    = 230 area.
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