Name: Area of Quadrilateral Using Heron's Formula
Uploaded: 2019-07-19
Duration: 10 min 24 s
Description: Did you know that the area of a quadrilateral can be calculated by dividing the quadrilateral into two triangles and using Heron's formula? Let's watch this...

Question

Find the area of a quadrilateral

A B C D

in which

A B = 3 c m, B C = 4 c m, C D = 4 c m, D A = 5 c m

and

A C = 5 c m .

Accepted Answer

In $Δ A B C$ , $AC$ is the hypotenuse as it is the longest side.

Let us check whether it is a right angle triangle or not using Pythagoras theorem.

$\Rightarrow A B^{2} + B C^{2}$

$\Rightarrow {(3)}^{2} + {(4)}^{2}$

$\Rightarrow 25$

$\Rightarrow {(5)}^{2}$

$\Rightarrow {AC}^{2}$

So, it satisfies Pythagoras theorem.

Hence, $Δ A B C$ is right-angled at $B$ .

So, the area of $ΔABC$ is,

$= [\frac{1}{2} \times A B \times B C]$

$= [\frac{1}{2} \times 3 \times 4] {cm}^{2}$

$= 6 {cm}^{2}$

For $Δ A C D$ ,

Semi perimeter, $s = \frac{(5 + 4 + 5)}{2} c m$

$= 7 cm$

According to Heron's formula, area of a triangle $= \sqrt{s (s - a) (s - b) (s - c)}$

Substituting the known values,

$= \sqrt{7 (7 - 5) (7 - 4) (7 - 5)} c m^{2}$

$= \sqrt{7 \times 2 \times 3 \times 2} c m^{2}$

$= 2 \sqrt{21} {cm}^{2}$

So, the area of $Δ A C D = (2 \times 4.583) c m^{2} = 9.166 c m^{2}$

Therefore, the area of the quadrilateral $A B C D =$ Area of $Δ A B C +$ Area of $Δ A C D$

$= (6 + 9.166) c m^{2}$

$= 15.166 c m^{2}$

Hence, the area of the quadrilateral $ABCD = 15.166 {cm}^{2}$ .

EXERCISE

Jharkhand Board Mathematics Solutions for EXERCISE

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