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
⇒ 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.
∴ 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.
THINGS TO REMEMBER
2. Area of a right angled triangle, whose base is b and perpendicular is p, is calculated by using the formula:
3. Area of an equilateral triangle, each of whose side is a, is calculated by using the formula:
4. Area of an isosceles triangle, each of whose equal sides is a and the unequal side is b, is calculated by using the formula:
|1. What is Heron's formula?|
|2. Can Heron's formula be used to find the area of a quadrilateral?|
|3. How do you split a quadrilateral into two triangles to use Heron's formula?|
|4. What is the advantage of using Heron's formula to find the area of a quadrilateral?|
|5. Is it possible to use Heron's formula to find the area of a quadrilateral when only the four sides are known?|