Application of Heron’s Formula in Finding Areas of Quadrilaterals

If we know the sides and one diagonal of the quadrilateral then we can find its area by using the Heron’s formula.

Find the area of the quadrilateral if its sides and the diagonal are given as follows.

Given, the sides of the quadrilateral

AB = 9 cm

BC = 40 cm

DC = 28 cm

AD = 15 cm

Diagonal is AC = 41 cm

Here, ∆ABC is a right angle triangle, so its area will be

Area of triangle =\dfrac{1}{2}\times\text{base}\times\text{height}

=\dfrac{1}{2}\times9\times40=180\ \text{cm}^2

Area of \triangle \text{ADC}=\sqrt{s(s-a)(s-b)(s-c)}

\mathrm{s=\dfrac{a+b+c}{2}}

=\dfrac{15+28+41}{2}=\dfrac{84}{2}=42

Area of ∆ADC = \sqrt{42(42-15)(42-28)(42-41)}=\sqrt{42(27)(14)(1)}=126\ \text{cm}^2

Area of Quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC

=180\ \text{cm}^2+126\ \text{cm}^2=306\ \text{cm}^2

Q.1: Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42cm.

Solution:

Assume that the third side of the triangle to be “x”.

Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm

It is given that the perimeter of the triangle = 42 cm

So, x = 42 – (18 + 10) cm = 14 cm

\therefore The semi perimeter of triangle =\dfrac{42}{2}=21\ \text{cm}

Using Heron’s formula,

Area of the triangle,

=\sqrt{[s(s-a)\ (s-b)\ (s-c)]}

=\sqrt{[21(21-18)\ (21-10)\ (21-14)]}\ \text{cm}^2

=\sqrt{[21\times3\times11\times7]}\ \text m^2

=21\sqrt{11}\ \text{cm}^2

Q.2: A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = ED = 10 m

AD = BE = 13 m

EC = 25 – ED = 25 – 10 = 15 m

Now, consider the triangle BEC,

Its semi perimeter (s) =\dfrac{(13+ 14 + 15)}{2}=21\ \text m

By using Heron’s formula,

Area of ΔBEC =

\sqrt{s(s-a)(s-b)(s-c)}

(\sqrt{21\times(21-13)\times(21-14)\times(21-15)})\ \text m^2

(\sqrt{21\times8\times7\times6})\ \text m^2

=84\ \text m^2

area of ΔBEC =\left(\dfrac{1}{2}\right)\times\text{CE}\times\text{BF}

84\ \text{cm}^2=\left(\dfrac{1}{2}\right)\times15\times \text{BF}

\Rightarrow BF=\left(\dfrac{168}{15}\right)\ \text{cm}=11.2\ \text{cm}

So, the total area of ABED will be BF × DE, i.e. 11.2 × 10 = 112 m2

\therefore Area of the field = 84 + 112 = 196 m2

Q.3: A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution:

Consider the triangle BCD,

Its semi-perimeter =  \dfrac{(48 + 30 + 30)}{2\ m}=54\ m

Using Heron’s formula,

Area of the ΔBCD =

\sqrt{s(s-a)(s-b)(s-c)}

 (\sqrt{54(54-48)(54-30)(54-30)})\ m^2

 (\sqrt{54\times6\times24\times24})\ m^2

=432\ m^2

\therefore Area of field = 2 × area of the ΔBCD = (2 × 432) m2 = 864 m2

Thus, the area of the grass field that each cow will be getting =  \left(\dfrac{864}{18}\right)\ \text m^2=48\ \text m^2

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