Question

A park in the shape of quadrilateral ABCD has \[\angle C = {90^ \circ }\], \[AB = 9\] m, \[BC = 12\] m, \[CD = 5\] m and \[AD = 8\] m. How much area does it occupy?

Answer 1

Hint:
We will find the area of the quadrilateral by dividing the quadrilateral into 2 triangles. We will find the area of each triangle using the formulas and add the areas to get the area of the quadrilateral.

Formulas used:
We will use the following formulas to solve the question.
1) Pythagoras Theorem: In a right-angled triangle, \[{H^2} = {P^2} + {B^2}\] where \[H,P\] and \[B\] are the hypotenuse, perpendicular and base of the triangle respectively.
2) Heron’s Formula: The area of a triangle with sides \[a,b\] and \[c\] is given by \[\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \] where \[s\] is the semi perimeter of the triangle and \[s = \dfrac{{a + b + c}}{2}\].
3) The area of a triangle is given by \[\dfrac{1}{2} \times b \times h\] where \[b\] and \[h\] are the base and height of the triangle respectively.

Complete step by step solution:
As \[\angle C = {90^ \circ }\], \[\Delta DBC\] is a right-angled triangle. We can apply Pythagoras theorem to it.
We will substitute 5 for P and 12 for B in the formula \[{H^2} = {P^2} + {B^2}\].
\[\begin{array}{c}{5^2} + {12^2} = {H^2}\\25 + 144 = {H^2}\end{array}\]
Adding the terms, we get
\[169 = {H^2}\]
Taking square root on both sides, we get
\[\begin{array}{c}\sqrt {169} = H\\13 = H\end{array}\]
We have found the hypotenuse BD of the triangle.
Let us find the area of \[\Delta DBC\]. We will substitute 12 for \[b\] and 5 for \[h\] in the formula \[\dfrac{1}{2} \times b \times h\].
\[{A_{DBC}} = \dfrac{1}{2} \times 12 \times 5 = 6 \times 5 = 30\]
Now, we will find the area of \[\Delta ABD\]. First, let us find the semi perimeter. We will substitute 9 for \[a\], 13 for \[b\] and 8 for \[c\]in the formula for \[s\].
\[\begin{array}{l}s = \dfrac{{9 + 13 + 8}}{2}\\s = \dfrac{{30}}{2}\\s = 15\end{array}\]
Now, let us find the semi perimeter. We will substitute 9 for \[a\], 13 for \[b\], 8 for \[c\]and 15 for \[s\] in the formula \[\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \].
\[\begin{array}{l}{A_{ABD}} = \sqrt {15\left( {15 - 9} \right)\left( {15 - 13} \right)\left( {15 - 8} \right)} \\{A_{ABD}} = \sqrt {15 \cdot 6 \cdot 2 \cdot 7} \\{A_{ABD}} = \sqrt {1260} \\{A_{ABD}} \approx 35.5\end{array}\]
Lastly, we will find the area of the quadrilateral by adding the areas of the 2 triangles.
\[\begin{array}{l}{A_{ABCD}} = 30 + 35.5 \\ {A_{ABCD}} = 65.5\end{array}\]

The area of the quadrilateral is 65.5 square units.

Note:
We have used Heron’s formula to find the area of one triangle because the sides are given, so we can easily find the perimeter and then the area. If we have used the formula for the area of the triangle then, we have to create a perpendicular line and then apply Pythagoras theorem to get the height. It would have become a much lengthier process. On the other hand we have used the formula of area of a triangle, because we can easily calculate the height using Pythagoras Theorem. Then using the value we can find the area.