Chapter 12 - Heron's Formula, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 PDF Download

INTRODUCTION

We are familiar with the shapes of many plane closed figures such as squares, rectangles, quadrilaterals, right triangles, equilateral triangles, isosceles triangles, scalene triangles, etc. We know the rules to find the perimeters and area of some of these figures. For example, a rectangle with length 12 m and width 8 m has perimeter equal to 2 (12 m + 8 m) = 40 m. The area of this rectangle is equal to (12 × 8) m² = 96 m². A square having each side of length 10 m has perimeter equal to 4 × 10 m = 40 m and area equal to 10² m² = 100 m².

Unit of measurement for length or breadth is taken as metre (m) or centimetre (cm) etc.

Unit of measurement for area of any plane figure is taken as square metre (m²) or square centimetre (cm²) etc.

In this section, we shall find the areas of some triangles.

AREA OF A TRIANGLE WITH GIVEN BASE AND HEIGHT

From your earlier classes, you know that:

Any side of the triangle may be taken as base and the length of perpendicular from the opposite vertex to the
base is the corresponding height.

For example, a triangle having base = 10 m and height = 6 m has its area equal to 30 m^2.

Area of a Right Triangle :– When the triangle is right angled, we can directly apply the above mentioned formula by using two sides containing the right angle as base and height.

In given figure, Area of  sq. units.

For example, a right angled triangle having two sides of length 3 m and 7 m (other than the hypotenuse), has its area

Area of an Equilateral Triangle :– Let ABC be an equilateral triangle with side a and AD be the perpendicular from A on BC. Then, D is the mid-point of BC i.e. BD = a/2

In right-angled ΔABD, by Pythagoras theorem, we have
AD² = AB² – BD²

So, area of 

Area of equilateral triangle with side a units 

For example, an equilateral triangle having side 8 cm, has its area

Area of an Isosceles Triangle :– Let ABC be an isosceles triangle with AB = AC = a and BC = b, and
AD be the perpendicular from A on BC.

Then, D is the mid-point of BC, i.e. BD = b/2

In right-angled ΔABD, by Pythagoras theorem we have :

So, area of  
 
Area of isosceles ΔABC with AB = AC = a units and BC = b units

For example, an isosceles triangle having equal sides of length 5 cm and unequal side of length 8 cm, has its area  

AREA OF A TRIANGLE BY USING HERON'S FORMULA
In a scalene triangle, if the length of each side is given but its height is not known and it cannot be obtained easily, we take the help of Heron's formula or Hero's formula given by Heron to find the area of such a triangle.
Heron's formula : If a,b,c denote the lengths of the sides of a triangle ABC. Then,

where  is the semi-perimeter of ΔABC.

Remark : This formula is applicable to all types of triangles whether it is right-angled or equilateral or isosceles.

Ex 1. Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm

Sol. Let a, b, c be the sides of the given triangle and s be its semi-perimeter such that a = 13 cm, b = 14 cm and c = 15 cm
Now,   = 21

 s – a = 21 – 13 = 8, s – b = 21 – 14 = 7 and s – c = 21 – 15 = 6

Hence, Area of given triangle 

= 7 × 4 × 3 = 84 cm²

^{Ex 2 . The perimeter of a triangular field is 450 m and its sides are in the ratio 13 : 12 : 5. Find the area of triangle.}

^{Sol. It is given that the sides a,b,c of the triangle are in the ratio 13 : 12 : 5 i.e.,
 a : b : c = 13 : 12 : 5 ⇒ a = 13x, b = 12x and c = 5x
 Perimeter = 450 ⇒ 13x + 12x + 5x = 450 ⇒ 30x = 450 ⇒ x = 15
 So, the sides of the triangle are
 a = 13 × 15 = 195 m, b = 12 × 15 = 180 m and c = 5 × 15 = 75 m
 It is given that perimeter = 450 ⇒ 2s = 450 ⇒ s = 225
 Hence,}  

^⇒  

^{⇒ Area}  = 6750 m²

Ex 3. Find the area of a triangle having perimeter 32 cm, one side 11 cm and difference of other two sides is 5 cm.

Sol. Let a, b and c be the three sides of ΔABC.
a = 11 cm
a + b + c = 32 cm ⇒ 11 + b + c = 32 cm or b + c = 21 cm ... (1)
Also, we are given that  b – c = 5 cm ... (2)
Adding (1) and (2), 2b = 26 cm
i.e., b = 13 cm and c = 8 cm
Now,   = 16 cm

(s – a) = (16 – 11) cm = 5 cm

(s – b) = (16 – 13) cm = 3 cm

(s – c) = (16 – 8) cm = 8 cm

= 

Ex 4. In figure, find the area of the ΔABC.

Sol.

Ex 5. The sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area.

Sol. Let us take the sides of the triangle as 3x, 5x and 7x because the ratio of the sides is given to be 3 : 5 : 7. Also, we are given that

Hence, the lengths of the three sides are 3 × 20 m, 5 × 20 m, 7 × 20 m. i.e., 60 m, 100 m, 140 m.
Now, 

Area of the triangle

Ex 6. The lengths of the sides of a triangle are 5 cm, 12 cm and 13 cm. Find the length of perpendicular from the opposite vertex to the side whose length is 13 cm.

Sol. Here, a = 5, b = 12 and c = 13.

Let p be the length of the perpendicular from vertex A on the side BC. Then,

Ex 7. In figure, there is a triangular childern park with sides, AB = 7 m, BC = 8 and AC = 5m, AD  BC and AD meets BC at D. Trees are planted at A, B, C and D. Find the distance between the trees at A and D.

Sol. In figure, a = 8 m, b = 5 m and c = 7 m.

The area of ΔABC

Ex 8. An isoscles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Sol. Area of an isosceles triangle  with equal side 'a' and base b.

Ex 9. Find the base of an isosceles triangle whose area is 12 cm2 and one equal sides is 5 cm.

Sol. Here equal sides : a = 5 cm, b = ?, Area = 12 cm²
Area of an isosceles triangle = 12 cm²


On squaring both sides, we get

Neglecting the negative sign as length cannot be –ve
 Base(b) = 8 cm or 6 cm

Ex 10. One side of an equilateral triangle measures 8 cm. Find the area using Heron's formula. What is its altitude?

Sol. Each side of an equilateral triangle = 8 cm
Here, a = 8 cm, b = 8 cm, c = 8 cm
  s – a = s – b = s – c = 12 – 8 = 4 cm