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In this page we have *NCERT book Solutions for Class 9th Maths:Heron Formula* for
EXERCISE 2 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Given in the question

∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

BD is joined.

Area of the quadrilateral ABCD can be found using the area of the separate triangle and then adding up

In ΔBCD,

By applying Pythagoras theorem,

BD

BD

BD

BD = 13 m

Area of ΔBCD=Area of right angle triangle= (1/2) × base × Height

So Area of ΔBCD = 1/2 × 12 × 5 = 30 m

Now,

Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m

Using heron's formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 - 13) (15 - 9) (15 - 8) m

= √15 × 2 × 6 × 7 m

= 6√35 m

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Given in the question

AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm

Area of the quadrilateral ABCD can be found using the area of the separate triangle and then adding up

In ΔABC,

By applying Pythagoras theorem,

AC

5

25 = 25

Thus, ΔABC is a right angled at B.

Area of ΔABC=Area of right angle triangle= (1/2) × base × Height

Area of ΔABC = 1/2 × 3 × 4 = 6 cm

Now,

Semi perimeter of ΔACD(s) = (5 + 5 + 4)/2 cm = 14/2 cm = 7 m

Using heron's formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √7(7 - 5) (7 - 5) (7 - 4) cm

= √7 × 2 × 2 × 3 cm

= 2√21 cm

Now Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm

Radha made a picture of an aeroplane with coloured paper as shown in below figure. Find the total area of the paper used.

Area of the aeroplane can be find using the area of the individual triangles and then summing them to get the total area

Length of the sides of the triangle section I = 5cm, 1cm and 5cm

Perimeter of the triangle = 5 + 5 + 1 = 11cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron's formula,

Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 - 5) (5.5 - 5) (5.5 - 1) cm

= √5.5 × 0.5 × 0.5 × 4.5 cm

= 0.75√11 cm

Length of the sides of the rectangle of section II = 6.5cm and 1cm

Area of section II = 6.5 × 1 cm

Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.

Area of the trapezium = 3 × √3/4 × 1

Section IV is right angled triangles with base 6cm and height 1.5cm

Area of region IV = 1/2 × 6 × 1.5cm

Section IV is right angled triangles with base 6cm and height 1.5cm

Area of region IV = 1/2 × 6 × 1.5cm

So Total area of the paper used = (2.488 + 6.5 + 1.3 + 4.5 + 4.5)cm

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Given,

Area of the parallelogram and triangle are equal.

Length of the sides of the triangle are 26 cm, 28 cm and 30 cm.

Perimeter of the triangle = 26 + 28 + 30 = 84 cm

Semi perimeter of the triangle = 84/2 cm = 42 cm

Using heron's formula,

Area of the triangle = √s (s-a) (s-b) (s-c)

= √42(42 - 26) (46 - 28) (46 - 30) cm

= √46 × 16 × 14 × 16 cm

= 336 cm

Now given

Area of parallelogram = Area of triangle

Base × h = 336 cm

28cm × h = 336 cm

h = 336/28 cm

h = 12 cm

The height of the parallelogram is 12 cm.

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

let PQRS be the rhombus shaped field and Diagonal QS divides the rhombus PQRS into two congruent triangles of equal area.

Semi perimeter of ΔPQS = (30 + 30 + 48)/2 m = 54 m

Using heron's formula,

Area of the ΔPQS = √s (s-a) (s-b) (s-c)

= √54(54 - 30) (54 - 30) (54 - 48) m

= √54 × 14 × 14 × 6 cm

= 432 m

Area of field = 2 × area of the ΔPQS = (2 × 432)m

So,

Area of grass field which each cow will be getting = 864/18 m

An umbrella is made by stitching 10 triangular pieces of cloth of two different colors as shown below in figure. Each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

The total Umbrella area is equal 5 times area of one triangle.

Semi perimeter of each triangular piece = (50 + 50 + 20)/2 cm = 120/2 cm = 60cm

Using heron's formula,

Area of the triangular piece = √s (s-a) (s-b) (s-c)

= √60(60 - 50) (60 - 50) (60 - 20) cm

= √60 × 10 × 10 × 40 cm

= 200√6 cm

Area of triangular piece = 5 × 200√6 cm

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in below. How much paper of each shade has been used in it?

We know from the property of square the diagonals of a square bisect each other at right angle.

Also Area of given kite = 1/2 (diagonal)

= 1/2 × 32 × 32 = 512 cm

Area of shade I = Area of shade II

So it is equal to

= 512/2 cm

So, area of paper required in each shade = 256 cm

For the III section, It is simple area calculation

Length of the sides of triangle = 6cm, 6cm and 8cm

Semi perimeter of triangle = (6 + 6 + 8)/2 cm = 10cm

Using heron's formula,

Area of the III triangular piece = √s (s-a) (s-b) (s-c)

= √10(10 - 6) (10 - 6) (10 - 8) cm

= √10 × 4 × 4 × 2 cm

= 8√6 cm

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm . Find the cost of polishing the tiles at the rate of 50p per cm

Semi perimeter of each triangular shape = (28 + 9 + 35)/2 cm = 36 cm

Using heron's formula,

Area of each triangular shape = √s (s-a) (s-b) (s-c)

= √36(36 - 28) (36 - 9) (36 - 35) cm

= √36 × 8 × 27 × 1 cm

= 36√6 cm

Total area of 16 tiles = 16 × 88.2 cm

Total cost of polishing the tiles = Rs. (1411.2 × 0.5) = Rs. 705.6

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10mand the non-parallel sides AD = 13m and BC = 14m.

We can drop a perpendicular from C on AM and draw a line CE parallel to DA

CM ⊥ AB and CE || AD.

This will divide the trapezium into two part

i) Parallelogram

ii) Triangle

Now

In ΔBCE,

BC = 14m, CE = AD = 13 m and

BE = AB - AE = 25 - 10 = 15m

Semi perimeter of the ΔBCE = (15 + 13 + 14)/2 m = 21 m

Using heron's formula,

Area of the ΔBCE = √s (s-a) (s-b) (s-c)

= √21(21 - 14) (21 - 13) (21 - 15) m

= √21 × 7 × 8 × 6 m

= 84 m

Now area of the ΔBCE = 1/2 × BE × CM = 84 m

1/2 × 15 × CM = 84 m

CM = 168/15 m

CM = 56/5 m

Area of the parallelogram AECD = Base × Altitude = AE × CM = 10 × 84/5 = 112 m

Area of the trapezium ABCD = Area of AECD + Area of ΔBCE

= (112+ 84) m

Download this assignment as pdf

Given below are the links of some of the reference books for class 9 Math.

- Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
- Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
- Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
- Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

Class 9 Maths Class 9 Science

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