Question

Given that sides of a quadrilateral are \[AB = 5m, {\text{ }}BC = 12m, {\text{ }}CD = 14m{\text{ and }}DA = 15{\text{ }}m\] respectively, and the angle contained by the first two sides is a right angle. Find its area.

Answer 1

Hint:
For solving this question, first of all, we will calculate the area of $\vartriangle ABC$, and by using the Pythagoras theorem we will calculate the side $AC$. Then we will calculate the perimeter of $\vartriangle ADC$ and then by using the heron’s formula we will calculate the area and at last, will add up both the areas.

Formula used:
Area of $\vartriangle ABC = \dfrac{1}{2} \times AB \times BC$
Here, $AB, BC$ are the sides of a triangle.
The perimeter of a triangle is equal to the sum of all the sides divided by two.
Heron’s formula for $\vartriangle ADC$
$ \Rightarrow \sqrt {S\left( {S - AD} \right)\left( {S - DC} \right)\left( {S - AC} \right)} $

Complete step by step solution:
We have the sides of a quadrilateral are given as \[AB = 5m,{\text{ }}BC = 12m,{\text{ }}CD = 14m{\text{ and }}DA = 15{\text{ }}m\]
So to calculate the area we will join $AC$,
The area of $\vartriangle ABC = \dfrac{1}{2} \times AB \times BC$
Putting the values, we get
$ \Rightarrow \dfrac{1}{2} \times 5 \times 12$
And on solving the above, we get
$ \Rightarrow 30c{m^2}$
Now $\vartriangle ABC$, by applying the Pythagoras theorem, we get
$ \Rightarrow A{C^2} = A{B^2} + B{C^2}$
Now on substituting the values, we get
$ \Rightarrow AC = \sqrt {{5^2} + {{12}^2}} $
On solving the above equation, we get
$ \Rightarrow AC = \sqrt {169} = 13m$
Now in $\vartriangle ADC$, let us assume $2S$ be the perimeter of the triangle, we get
$ \Rightarrow 2S = AD + DC + AC$
So on putting the values, it can be written as
$ \Rightarrow S = \dfrac{1}{2}\left( {15 + 14 + 13} \right)$
And on solving more, we get
$ \Rightarrow S = \dfrac{1}{2} \times 42 = 21m$
Now by using the heron’s formula, we get
The area $\vartriangle ADC$ will be given as
$ \Rightarrow \sqrt {S\left( {S - AD} \right)\left( {S - DC} \right)\left( {S - AC} \right)} $
Putting the values, we get
$ \Rightarrow \sqrt {21\left( {21 - 15} \right)\left( {21 - 14} \right)\left( {21 - 13} \right)} $
And on solving the above equation, we get
$ \Rightarrow \sqrt {21 \times 6 \times 7 \times 8} = 84{m^2}$
Therefore, the area of quadrilateral $ABCD$ will be
$ \Rightarrow Area{\text{ of }}\vartriangle {\text{ABC + Area of }}\vartriangle {\text{ADC}}$
So on putting the values, we get
$ \Rightarrow \left( {30 + 84} \right){m^2} = 114{m^2}$

Therefore, the $114{m^2}$ will be the area of a quadrilateral.

Note:
For solving this type of problem we should always go through the diagram because it makes it easier to find the result. And the next thing is we should have to remember some formulas which we will use everywhere. Through these two we can solve such problems easily.