Finding Area of Quadrilaterals using Herons Formula - Herons Formula, Class 9, Mathema | Extra Documents & Tests for Class 9 PDF Download

APPLICATIONS OF HERON'S FORMULA IN FINDING AREA OF A QUADRILATERAL

Heron's formula can be applied to find the area of a quadrilateral by dividing the quadrilateral into two triangular parts. If we join any of the two diagonals of the quadrilateral, then we get two triangles. Area of each triangle is calculated and the sum of two areas is the area of the quadrilateral.

1. RHOMBUS : A quadrilateral whose all the sides are equal and opposite sides are parallel. The diagonals of a rhombus bisect each other at right angle.
In figure, ABCD is a rhombus with side a and AC and BD are the two diagonals and AC = d₁, BD = d₂, then

Perimeter = 4a

Area = Area of ΔABD + Area of ΔCDB

= 1/2. BD . OA + 1/2 . BD . OC
= 1/2.BD[OA + OC] = 1/2.BD . AC = 1/2. × d₁ × d₂ or 1/2 . × product of diagonals.
 

2. TRAPEZIUM : A quadrilateral which has one pair of opposite sides parallel.

In figure, ABCD is a trapezium with AB || CD,

AB = b, BC = c, CD = a, DA = d and the height of the trapezium ABCD is h, then

Perimeter = a + b + c + d

Area = Area of ΔAMD + Area of rectangle MNCD + Area of  ΔCNB
= 1/2. y . h + a . h + 1/2 . x . h
= 1/2. h (x + y) + ah = 1/2. h [x + y + 2a]
= 1/2. h [(x + y + a) + a] = 1/2. h [b + a]
= 1/2. (a + b) h = 1/2. (Sum of the parallel sides) × Distance between them.

3. QUADRILATERAL : A figure, which has four sides and four vertices. Generally a quadrilateral is denoted bythe symbol 'rectangle'.

In figure, ABCD is a quadrilateral in which AC is a diagonal and AC divides quadrilateral ABCD in two triangle ADC and ABC. Also, h₁ and h₂ are the altitudes of the triangle ADC and ABC respectively.
∴ Area = Area of ΔADC + Area of ΔABC

=1/2 . AC . h₁ + 1/2. AC . h₂ = 1/2. AC (h₁ + h₂)
=  1/2 . × One diagonal × (Sum of heights of those triangles whose base is former diagonal)

Ex. Find the area of the quadrilateral ABCD, in which AB = 7 cm, BC = 6 cm, CD= 12 cm, DA = 15 cm and AC = 9 cm.

Sol.The diagonal AC divides the quadrilateral ABCD into two triangles ABC and ACD.

Area of quad. ABCD = Area of AABC + Area of ΔACD, For ΔABC, we have

Semiperuneter 

⇒ A₁ = Area of ΔABC
⇒ At = Area of ΔABC =

⇒ At = Area of ΔABC = 20.98 cm²
For ΔACD, we have

⇒ A₂ = Area of ΔACD   = 9 × 2 × 3 = 54 cm²
Hence, Area of quad. ABCD = A₁ + A₂ = (20.98 + 54) cm₂ = 74.98 cm²

Ex. Prove that the area of the quadrilateral ABCD is  if AB = 5 m, BC = 5 m, CD = 6 m,AD = 6 m, and diagonal AC = 6 m.

Sol. Diagonal AC divides the quadrilateral ABCD into two triangles ΔACD and ΔABC.
For ΔACD, side are 6m, 6 m and 6 m.

Semiperimeter,
∴ Area of ΔACD =  m²

For ΔABC, side are 5 m, 5 m and 6 m.
Semiperimeter,  = 8m

Area of ΔABC =

Thus, the area of the quadrilateral ABCD =

Ex. In fig. ABCD is a field in the form of a quadrilateral whose sides are indicated in the figure. lf DAB = 90°, find the area of the field.

Sol. Clearly, ΔDAB is a right-angled triangle. Therefore,
DB² = DA² + AB² [Using Pythagoras Theorem]

Ex. A rhombus has perimeter 100 m and one of its diagonal is 40 m. Find the area of the rhombus.

Sol. ABCD is the rhombus having perimeter = 100 m and AC = 40 m. 

Now, we have AB = BC = CD = AD =100/4 m = 25 m

We know that, ar (ΔABC) = ar (ΔADC)

Sides of ΔABC are 25 m, 25 m and 40 m.
Semi perimeter of ΔABC (s)
The area of ΔABC
 = 3 × 5 × 20 m² = 300 m²
Also, we havearea of ΔADC = 300 m².
Hence, the area of the rhombus ABCD = ar (ΔABC) + ar (ΔADC) = 300 m² + 300 m² = 600 m².