Class 9 Application of Heron’s Formula in Finding Areas of Quadrilaterals

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Class 9 chapter - 12 HERON’S FORMULA

Application of Heron’s Formula in Finding Areas of Quadrilaterals

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♦ Application of Heron’s Formula in Finding Areas of Quadrilaterals

Application of Heron’s Formula in Finding Areas of Quadrilaterals]

Suppose that a farmer has a land to be cultivated and she employs some labourers for this purpose on the terms of wages calculated by area cultivated per square metre.

How will she do this? Many a time, the fields are in the shape of quadrilaterals. We need to divide the quadrilateral in triangular parts and then use the formula for area of the triangle.

Q 3210578410

Kamla has a triangular field with sides` 240 m, 200 m, 360 m`, where she
grew wheat. In another triangular field with sides `240 m, 320 m, 400 m` adjacent to the
previous field, she wanted to grow potatoes and onions (see Fig. 12.11). She divided
the field in two parts by joining the mid-point of the longest side to the opposite vertex
and grew patatoes in one part and onions in the other part. How much area (in hectares)
has been used for wheat, potatoes and onions? (1 hectare `= 10000 m^2`)
Class 9 Chapter 12 Example 4

Solution:

Let `ABC` be the field where wheat is grown. Also let `ACD` be the field
which has been divided in two parts by joining C to the mid-point E of AD. For the
area of triangle `ABC`, we have

`a= 200 m , b = 240 m , c= 360 m`

Therefore, `s = ( 200 +240 +360)/2 m = 400 m` .

So, area for growing wheat

` = sqrt ( 400 (400 - 200 )( 400 -240 )( 400 - 360 ) ) m^2 `

`= sqrt ( 400 xx 200 xx 160 xx 40 ) m^2 `

`= 16000 sqrt 2 m^2 = 1.6 xx sqrt 2 ` hectares

` = 2.26 ` hectares (nearly)

Let us now calculate the area of triangle `ACD`.

Here, we have `s = (240 + 320 + 400)/2 m = 480 m` .

So, area of `Delta ACD = sqrt (480 (480- 240)( 480 - 320 )( 480 -400) ) m^2`

`= sqrt (480 xx 240 xx 160 xx 80) m^2 = 38400 m^2 = 3.84 ` hectares

We notice that the line segment joining the mid-point `E` of `AD` to `C` divides the
triangle `ACD` in two parts equal in area. Can you give the reason for this? In fact, they
have the bases `AE` and `ED` equal and, of course, they have the same height.

Therefore, area for growing potatoes = area for growing onions

`= (3.84 ÷ 2)` hectares `= 1.92` hectares.

Q 3220578411

Students of a school staged a rally for cleanliness campaign. They
walked through the lanes in two groups. One group walked through the lanes `AB, BC`
and `CA`; while the other through `AC, CD` and `DA` (see Fig. 12.12). Then they cleaned
the area enclosed within their lanes. If `AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m`
and `angle B = 90º`, which group cleaned more area and by how much? Find the total area
cleaned by the students (neglecting the width of the lanes).
Class 9 Chapter 12 Example 5

Solution:

Since `AB = 9 m and BC = 40 m, angle B = 90°`, we have:

`AC = sqrt ( 9^2 + 40^2 ) m`

`= sqrt (81 + 1600) m `

` = sqrt (1681) m = 41 m`

Therefore, the first group has to clean the area of triangle `ABC`, which is right angled.

Area of `Delta ABC = 1/2 xx text (base) xx text (height)`

`= 1/2 xx 40 xx 9 m^2 = 180 m^2`

The second group has to clean the area of triangle `ACD`, which is scalene having sides
`41 m, 15 m` and `28 m`.

Here, `s = (41 +15 +28)/2 m = 42 m`

Therefore, area of `Delta ACD = sqrt ( s (s-a)(s-b)(s-c) )`

`= sqrt ( 42 (42- 41)( 42 -15)( 42 -28)) m^2`

`= sqrt (42 xx 1 xx 27 xx 14 ) m^2 = 126 m^2`

So first group cleaned `180 m^2` which is `(180 – 126) m^2`, i.e., `54 m^2` more than the area
cleaned by the second group.

Total area cleaned by all the students `= (180 + 126) m^2 = 306 m^2`.

Q 3240578413

Sanya has a piece of land which is in the shape of a rhombus
(see Fig. 12.13). She wants her one daughter and one son to work on the land and
produce different crops. She divided the land in two equal parts. If the perimeter of
the land is 400 m and one of the diagonals is` 160 m`, how much area each of them will
get for their crops?
Class 9 Chapter 12 Example 6

Solution:

Let `ABCD` be the field.

Perimeter `= 400 m`

So, each side `= 400 m ÷ 4 = 100 m`

i.e. `AB = AD = 100 m`.

Let diagonal `BD = 160 m`.

Then semi-perimeter `s` of `Delt ABD` is given by

`s = (100 + 100 + 160)/2 m = 180 m`

Therefore, area of `Delta ABD = sqrt ( 180 (180-100)(180-100)(180-160) )`

`= sqrt ( 180 xx 80 xx 80 xx 20) m^2 = 4800 m^2`

Therefore, each of them will get an area of `4800 m^2`.

`text (Alternative method :)` Draw `CE ⊥ BD `(see Fig. 12.14).

As `BD = 160 m`, we have

`DE = 160 m ÷ 2 = 80 m`

And, `DE^2 + CE^2 = DC^2`, which gives

`CE = sqrt (DC^2 -DE^2)`

or, `CE = sqrt (100^2 - 80^2) m = 60 m`

Therefore, area of `Delta BCD = 1/2 xx 160 xx 60 m^2 = 4800 m^2`