Please Wait... While Loading Full Video#### Class 9 chapter - 12 HERON’S FORMULA

♦ Introduction

♦ Area of a Triangle — by Heron’s Formula

♦ Area of a Triangle — by Heron’s Formula

You have studied in earlier classes about figures of different shapes such as squares, rectangles, triangles and quadrilaterals.

You have also calculated perimeters and the areas of some of these figures like rectangle, square etc. For instance, you can find the area and the perimeter of the floor of your classroom.

Let us take a walk around the floor along its sides once; the distance we walk is its perimeter. The size of the floor of the room is its area.

So, if your classroom is rectangular with length `10 m` and width `8 m`, its perimeter would be `2(10 m + 8 m) = 36 m` and its area would be `10 m × 8 m`, i.e., `80 m^2`.

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 `(m^2)` or square centimetre `(cm^2)` etc.

Suppose that you are sitting in a triangular garden. How would you find its area? From Chapter 9 and from your earlier classes, you know that:

We see that when the triangle is `text ( right angled )`, we can directly apply the formula by using two sides containing the right angle as base and height.

For example, suppose that the sides of a right triangle `ABC` are `5 cm, 12 cm `and `13 cm`; we take base as `12 cm` and height as `5 cm` (see Fig. 12.1). Then the

area of `Delta ABC` is given by

`1/2 xx text (base) xx text(height) = 1/2 xx 12 xx 5 cm^2` , i.e., `30 cm^2`

Note that we could also take `5 cm` as the base and` 12 cm` as height.

Now suppose we want to find the area of an equilateral triangle `PQR` with side `10cm` (see Fig. 12.2). To find its area we need its height. Can you find the height of this triangle?

Let us recall how we find its height when we know its sides. This is possible in an equilateral triangle. Take the mid-point of `QR` as `M` and join it to `P`.

We know that `PMQ` is a right triangle. Therefore, by using Pythagoras Theorem, we can find the length `PM` as shown below:

`PQ^2 = PM^2 + QM^2`

i.e., `(10)^2 = PM^2 + (5)^2`, since `QM = MR`.

Therefore, we have `PM^2 = 75`

i.e., `PM = sqrt (75) cm = 5 sqrt (3) cm` .

Then area of `Delta PQR = 1/2 xx text(base) xx text (height) = 1/2 xx 10 xx 5 sqrt 3 cm^2 = 25 sqrt 3 cm^2` .

Let us see now whether we can calculate the area of an`text ( isosceles triangle )` also with the help of this formula. For example, we take a triangle `XYZ` with two equal sides `XY` and `XZ` as `5 cm` each and unequal side `YZ` as `8 cm` (see Fig. 12.3).

In this case also, we want to know the height of the triangle. So, from `X` we draw a perpendicular `XP` to side `YZ`. You can see that this perpendicular `XP` divides the base `YZ` of the triangle in two equal parts.

Therefore, `YP = PZ = 1/2 YZ =4 cm`

Then, by using Pythagoras theorem, we get

`XP^2 = XY^2 – YP^2`

`= 5^2 – 4^2 = 25 – 16 = 9`

So, `XP = 3 cm`

Now, area of `Delta XYZ = 1/2 xx text (base) YZ xx text (height) XP`

`= 1/2 xx 8 xx 3 cm^2 = 12 cm^2`

Now suppose that we know the lengths of the sides of a scalene triangle and not the height. Can you still find its area? For instance, you have a triangular park whose sides are `40 m, 32 m`, and `24 m`.

How will you calculate its area? Definitely if you want to apply the formula, you will have to calculate its height. But we do not have a clue to calculate the height. Try doing so. If you are not able to get it, then go to the next section.

You have also calculated perimeters and the areas of some of these figures like rectangle, square etc. For instance, you can find the area and the perimeter of the floor of your classroom.

Let us take a walk around the floor along its sides once; the distance we walk is its perimeter. The size of the floor of the room is its area.

So, if your classroom is rectangular with length `10 m` and width `8 m`, its perimeter would be `2(10 m + 8 m) = 36 m` and its area would be `10 m × 8 m`, i.e., `80 m^2`.

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 `(m^2)` or square centimetre `(cm^2)` etc.

Suppose that you are sitting in a triangular garden. How would you find its area? From Chapter 9 and from your earlier classes, you know that:

`text(Area of a triangle ) =1/2 xx text(base) xx text (height) `

(I)We see that when the triangle is `text ( right angled )`, we can directly apply the formula by using two sides containing the right angle as base and height.

For example, suppose that the sides of a right triangle `ABC` are `5 cm, 12 cm `and `13 cm`; we take base as `12 cm` and height as `5 cm` (see Fig. 12.1). Then the

area of `Delta ABC` is given by

`1/2 xx text (base) xx text(height) = 1/2 xx 12 xx 5 cm^2` , i.e., `30 cm^2`

Note that we could also take `5 cm` as the base and` 12 cm` as height.

Now suppose we want to find the area of an equilateral triangle `PQR` with side `10cm` (see Fig. 12.2). To find its area we need its height. Can you find the height of this triangle?

Let us recall how we find its height when we know its sides. This is possible in an equilateral triangle. Take the mid-point of `QR` as `M` and join it to `P`.

We know that `PMQ` is a right triangle. Therefore, by using Pythagoras Theorem, we can find the length `PM` as shown below:

`PQ^2 = PM^2 + QM^2`

i.e., `(10)^2 = PM^2 + (5)^2`, since `QM = MR`.

Therefore, we have `PM^2 = 75`

i.e., `PM = sqrt (75) cm = 5 sqrt (3) cm` .

Then area of `Delta PQR = 1/2 xx text(base) xx text (height) = 1/2 xx 10 xx 5 sqrt 3 cm^2 = 25 sqrt 3 cm^2` .

Let us see now whether we can calculate the area of an`text ( isosceles triangle )` also with the help of this formula. For example, we take a triangle `XYZ` with two equal sides `XY` and `XZ` as `5 cm` each and unequal side `YZ` as `8 cm` (see Fig. 12.3).

In this case also, we want to know the height of the triangle. So, from `X` we draw a perpendicular `XP` to side `YZ`. You can see that this perpendicular `XP` divides the base `YZ` of the triangle in two equal parts.

Therefore, `YP = PZ = 1/2 YZ =4 cm`

Then, by using Pythagoras theorem, we get

`XP^2 = XY^2 – YP^2`

`= 5^2 – 4^2 = 25 – 16 = 9`

So, `XP = 3 cm`

Now, area of `Delta XYZ = 1/2 xx text (base) YZ xx text (height) XP`

`= 1/2 xx 8 xx 3 cm^2 = 12 cm^2`

Now suppose that we know the lengths of the sides of a scalene triangle and not the height. Can you still find its area? For instance, you have a triangular park whose sides are `40 m, 32 m`, and `24 m`.

How will you calculate its area? Definitely if you want to apply the formula, you will have to calculate its height. But we do not have a clue to calculate the height. Try doing so. If you are not able to get it, then go to the next section.

The formula given by Heron about the area of a triangle, is also known as Hero’s formula. It is stated as:

where `a, b` and` c` are the sides of the triangle, and `s = text (semi-perimeter )` , i.e., half the

This formula is helpful where it is not possible to find the height of the triangle easily. Let us apply it to calculate the area of the triangular park` ABC`, mentioned above (see Fig. 12.5).

Let us take `a = 40 m, b = 24 m, c = 32 m`,

so that we have` s = (40+24+32)/2 m = 48 m` .

`s – a = (48 – 40) m = 8 m`,

`s – b = (48 – 24) m = 24 m`,

`s – c = (48 – 32) m = 16 m`.

Therefore, area of the park `ABC`

` = sqrt ( s (s-a) (s-b) (s-c) )`

`= sqrt (48 xx 8 xx 24 xx 16 ) m^2 = 384 m^2`

We see that `32^2 + 24^2 = 1024 + 576 = 1600 = 40^2`. Therefore, the sides of the park make a right triangle. The largest side, i.e., BC which is `40 m` will be the hypotenuse and the angle between the sides `AB` and `AC` will be `90°`.

By using Formula I, we can check that the area of the park is `1/2 xx 32 xx 24m^2 = 384 m^2` .

We find that the area we have got is the same as we found by using Heron’s formula.

Now using Heron’s formula, you verify this fact by finding the areas of other triangles discussed earlier viz.,

(i) equilateral triangle with side `10 cm`.

(ii) isosceles triangle with unequal side as `8 cm` and each equal side as `5 cm`. You will see that

For (i), we have `s = (10+10+10)/2 cm = 15 cm` .

`text (Area of triangle ) = sqrt (15 (15-10) (15-10) (15-10) ) cm^2 `

`= sqrt (15 xx 5 xx 5 xx 5) cm^2 = 25 sqrt 3 cm^2`

For (ii), we have `s = ( 8+5 +5)/2 cm = 9 cm`

`text (Area of triangle ) = sqrt ( 9 (9-8) (9-5) (9-5) ) cm^2 = sqrt ( 9 xx 1 xx 4 xx 4) cm^2 = 12 cm^2` .

Let us now solve some more examples:

`text (Area of a triangle ) = sqrt ( s (s-a) (s-b) (s-c) )` ..............(II)

where `a, b` and` c` are the sides of the triangle, and `s = text (semi-perimeter )` , i.e., half the

Perimeter of the triangle `= (a+b +c)/2` ,

This formula is helpful where it is not possible to find the height of the triangle easily. Let us apply it to calculate the area of the triangular park` ABC`, mentioned above (see Fig. 12.5).

Let us take `a = 40 m, b = 24 m, c = 32 m`,

so that we have` s = (40+24+32)/2 m = 48 m` .

`s – a = (48 – 40) m = 8 m`,

`s – b = (48 – 24) m = 24 m`,

`s – c = (48 – 32) m = 16 m`.

Therefore, area of the park `ABC`

` = sqrt ( s (s-a) (s-b) (s-c) )`

`= sqrt (48 xx 8 xx 24 xx 16 ) m^2 = 384 m^2`

We see that `32^2 + 24^2 = 1024 + 576 = 1600 = 40^2`. Therefore, the sides of the park make a right triangle. The largest side, i.e., BC which is `40 m` will be the hypotenuse and the angle between the sides `AB` and `AC` will be `90°`.

By using Formula I, we can check that the area of the park is `1/2 xx 32 xx 24m^2 = 384 m^2` .

We find that the area we have got is the same as we found by using Heron’s formula.

Now using Heron’s formula, you verify this fact by finding the areas of other triangles discussed earlier viz.,

(i) equilateral triangle with side `10 cm`.

(ii) isosceles triangle with unequal side as `8 cm` and each equal side as `5 cm`. You will see that

For (i), we have `s = (10+10+10)/2 cm = 15 cm` .

`text (Area of triangle ) = sqrt (15 (15-10) (15-10) (15-10) ) cm^2 `

`= sqrt (15 xx 5 xx 5 xx 5) cm^2 = 25 sqrt 3 cm^2`

For (ii), we have `s = ( 8+5 +5)/2 cm = 9 cm`

`text (Area of triangle ) = sqrt ( 9 (9-8) (9-5) (9-5) ) cm^2 = sqrt ( 9 xx 1 xx 4 xx 4) cm^2 = 12 cm^2` .

Let us now solve some more examples:

Q 3270478316

Find the area of a triangle, two sides of which are `8 cm` and `11 cm` and

the perimeter is `32 cm` (see Fig. 12.6).

Class 9 Chapter 12 Example 1

the perimeter is `32 cm` (see Fig. 12.6).

Class 9 Chapter 12 Example 1

Here we have perimeter of the triangle` = 32 cm, a = 8 cm` and` b = 11 cm`.

Third side` c = 32 cm – (8 + 11) cm = 13 cm`

So,`2s = 32,` i.e., `s = 16 cm`,

`s – a = (16 – 8) cm = 8 cm`,

`s – b = (16 – 11) cm = 5 cm`,

`s – c = (16 – 13) cm = 3 cm.`

Therefore, area of the triangle = ` sqrt ( s (s-a)(s-b)(s-c))`

`= sqrt (16 xx 8 xx 5 xx 3) cm^2 = 8 sqrt (30 ) cm^3`

Q 3280478317

A triangular park `ABC` has sides` 120m, 80m` and` 50m` (see Fig. 12.7). A

gardener Dhania has to put a fence all around it and also plant grass inside. How

much area does she need to plant? Find the cost of fencing it with barbed wire at the

rate of Rs `20` per metre leaving a space `3m` wide for a gate on one side.

Class 9 Chapter 12 Example 2

gardener Dhania has to put a fence all around it and also plant grass inside. How

much area does she need to plant? Find the cost of fencing it with barbed wire at the

rate of Rs `20` per metre leaving a space `3m` wide for a gate on one side.

Class 9 Chapter 12 Example 2

For finding area of the park, we have

`2s = 50 m + 80 m + 120 m = 250 m`.

i.e., `s = 125 m`

Now, `s – a = (125 – 120) m = 5 m`,

`s – b = (125 – 80) m = 45 m`,

`s – c = (125 – 50) m = 75 m`

Therefore, area of the park` = sqrt ( s(s-a)(s-b)(s-c) )`

`= sqrt (125 xx 5 xx 45 xx 75) m^2 `

`= 375 sqrt (15) m^2`

Also, perimeter of the park `= AB + BC + CA = 250 m`

Therefore, length of the wire needed for fencing `= 250 m – 3 m` (to be left for gate)

`= 247 m`

And so the cost of fencing = Rs` 20 × 247 =` Rs `4940`

Q 3200478318

The sides of a triangular plot are in the ratio of `3 : 5 : 7` and its perimeter

is `300 m`. Find its area.

Class 9 Chapter 12 Example 3

is `300 m`. Find its area.

Class 9 Chapter 12 Example 3

Suppose that the sides, in metres, are `3x, 5x` and `7x` (see Fig. 12.8).

Then, we know that `3x + 5x + 7x = 300` (perimeter of the triangle)

Therefore, `15x = 300`, which gives `x = 20`.

So the sides of the triangle are `3 × 20 m, 5 × 20 m `and `7 × 20 m`

i.e.,` 60 m, 100 m` and `140 m`.

Can you now find the area [Using Heron’s formula]?

We have `s = (60+ 100 +140 )/2 m = 150 m` ,

and area will be `sqrt ( 150 (150 -60) (150 -100) (150 -140) )m^2`

`= sqrt (150 xx 90 xx 50 xx 10 ) m^2`

`= 1500 sqrt 3 m^2`