Class 8 – Mensuration

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Topic Sub Topic Online Practice Test
  • Perimeter and area of quadrilaterals
  • Perimeter and Area of compound figures
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  • Circumference and area of a circle, semi-circle
  • Perimeter and Area of paths
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Introduction: We have learnt that for a closed plane figure, the perimeter is the measure around its boundary and its area is the region covered by it.
We found the area and perimeter of various plane figures such as triangles, rectangles etc…. Let us recall them and move further to find area and perimeter of polygons like quadrilaterals, trapezium, hexagon etc…..

Perimeter and area of basic shapes.


Perimeter and area of an equilateral triangle:
Let us take an equilateral triangle ABC with each side measuring a units.

Area = BC × AD

By Pythagoras theorem

Heron’s formula: Area of triangle with sides  is given by

where  semi perimeter.

Area of a quadrilateral: Let ABCD be a quadrilateral with AC as one of its diagonal. Let BP and DQ be the perpendiculars drawn from the vertices B and D to the diagonal AC.

From the figure,

Area of quadrilateral ABCD


Where d denotes length of diagonal AC.

Diagonal × sum of perpendiculars on the diagonal from the opposite vertices.














l × b




2(Sum of the adjacent sides)


b × h

Area of rhombus: Let ABCD be a rhombus and ACand BD be its diagonals which intersect at O. We know that the diagonals of a rhombus bisect each other at right angles and divides the rhombus into four congruent right triangles.

Area of rhombus = 4 × Area of


Area of rhombus Productof its diagonals of rhombus .


Area of trapezium: A trapezium is a quadrilateral having a pair of parallel opposite sides. Let ABCD be a trapezium in which units, where h is the height of trapezium.

Area of trapezium ABCD = Area of




Hence, area of trapezium  (sum of parallel sides)  (height)

Area of polygon: We can calculate the area of irregular polygon by dividing them into triangles or quadrilaterals or a combination of the two.

Suppose we have to find the area of polygon MNOPQR.

We can find out into 2 ways.

i)     By dividing the polygon MNOPQR
into two trapeziums.

ii)   By dividing the polygon MNOPQR intoMNO, Quadrilateral MOPR andPQR

Now let us learn, how to find the area of a regular polygon.

Area of a regular polygon: The regular polygon ofn sides may be divided into n triangles of equal areas. Consider the following regular octagon.

The octagon can be divided into 8 equal triangles. Let OP be the
perpendicular distance from centre O to side AH. OP is the height of the triangle OHA.

Area of triangle  sq.units

Since the octagon consists of 8 equal triangles.

Area of the octagon

But the perimeter of the octagon = 8(AH) units

Hereis a formula which relates the perimeter of a regular polygon to its area.

 The area of a regular polygon (Perimeter of the polygon) × (Perpendicular distance from the centre to any side.)

1.   The sides of quadrilateral taken in order are 8 m, 8 m, 7 m and 5 m respectively and the angle contained by the first two sides is 60°, find its perimeter and area.

Solution: Area of quad

We shall use heron’s formula for the triangles BDC.

ΔABDis an equilateral Δle, so BD = 8 cm

2.   The diagonals of a rhombus are 24 cm and 10 cm. What is the perimeter of the rhombus?

Solution: Each side =

 Perimeter= .

3.   The area of a rhombus is 150 cm2. The length of one of its diagonals is 10 cm. What is the length of the other diagonal?

Solution: Let the length of second diagonal = cm

Then, .

4.   The area of a field in the shape of a trapezium measures 1440 m2. The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3 , what is the length of the longer parallel side?

Solution: Let the parallel sides be 5and 3x respectively.


The longer parallel side = .

5.   The length of the floor of a rectangular hall is 10 m more than its breadth.If 34 carpets of size 6 m × 4 m are required to cover the floor of the hall, then find the length and breadth of the hall.

Solution: Let length and breadth of rectangular hall be and b.

It isgiven l = (b+ 10)

Area of 34 carpets = Area of floor of hall

34 × 6× 4 = b × (b+10)

 b× (b +10) = 24 × (24 + 10)    

Fromabove b = 24 m

 =24 + 10 = 34 m

∴Length of rectangular hall = 34m

   Breadth of rectangular hall = 24m

6.   Findthe area of the pentagon ABCDE shown below if AD = 8 cm, AH = 6 cm, AG = 4 cm,AF = 3 cm, BF = 2 cm, CH = 3 cm and EG = 2.5cm.

Solution: Area of pentagon ABCDE = Areaof ΔAFB + Area of trapezium BCHF + Area of ΔCHD + Area of ΔADE.

∴ Areaof Pentagon = 23.5 cm2

Given figure shows a regular hexagon ABCDEF of side 5 cm.Find its area in two different ways.



Method – I :  We join BE, which will divide the hexagon into two trapeziums of equal areas.

∴ Area of hexagon ABCDEF = Area of Trapezium ABEF +Area of Trapezium BCDE

= 2 × (Area of Trapezium ABEF)

For FG use Pythagoras theorem.

Area of hexagon ABCDEF =

Area of hexagon ABCDEF = 64 cm2

Method – II: Here we join AC and FD. The hexagon is now divided into two triangles ABC and FED of equal areas and a rectangle ACDF

Area of hexagon ABCDEF = 2 × Area ΔABC+Area of rectangle ACDF