Question

In a quadrilateral $ABCD$, if \[AB = 7cm,{\text{ }}BC = 12cm,{\text{ }}CD = 12{\text{ }}cm,{\text{ }}DA = 9{\text{ }}cm\]and diagonal \[AC = 15\]cm. What is its area?
A) $(10\sqrt {17} + 54)$ sq. cm
B) $(10\sqrt {34} - 54)$sq. cm
C) Data insufficient
D) None of these

Answer 1

Hint:Break the given quadrilateral in two triangles and then find the area of the quadrilateral by finding the sum of the areas of these triangles. We can find the area of the triangles using Heron’s formula of the area of the triangle.

Complete step-by-step answer:
It is given that \[AB = 7cm,{\text{ }}BC = 12cm,{\text{ }}CD = 12{\text{ }}cm,{\text{ }}DA = 9{\text{ }}cm\]and diagonal \[AC = 15\]cm in a quadrilateral $ABCD$.
We have to find the area of the quadrilateral.
Sketch the figure of the quadrilateral $ABCD$ using the given data:

We can divide the quadrilateral into two triangles: triangle $ADC$ and triangle $ABC$.
So, the area of the quadrilateral can be obtained by finding the area of these two triangles.
Then the area of the quadrilateral is given as:
Area of Quadrilateral $ABCD$$ = {\text{Area of triangle ABC + Area of triangle ADC}}$
First, find the area of the triangles $ABC$ and $ADC$.
In the triangle $ABC$,
$AB\left( a \right) = 7cm,BC\left( b \right) = 12cm$ and $AC\left( c \right) = 15cm$,
\[s = \dfrac{{a + b + c}}{2}\]
Substitute the value of $a,b$ and $c$ into the formula:
$s = \dfrac{{7 + 12 + 15}}{2}$
$s = \dfrac{{34}}{2} = 17$
Now, use the Heron’s formula to find the area of the triangle, which is given as:
Area of triangle $ABC$$ = \sqrt {s(s - a)(s - b)(s - c)} $
Substitute the values of $s,a,b,$ and $c$ into the formula:
Area of triangle $ABC$$ = \sqrt {17(17 - 7)(17 - 12)(17 - 15)} $
Area of triangle $ABC$$ = \sqrt {17 \times 10 \times 5 \times 2} $
Area of triangle $ABC$$ = \sqrt {1700} $
Area of triangle $ABC$$ = 10\sqrt {17} {\text{ c}}{{\text{m}}^2}$
So, the area of the triangle $ABC$ is $10\sqrt {17} {\text{ c}}{{\text{m}}^2}$
In the triangle $ADC$,
$AD\left( a \right) = 9cm,DC\left( b \right) = 12cm$and$AC\left( c \right) = 15cm$,
\[s = \dfrac{{a + b + c}}{2}\]
Substitute the value of $a,b$ and $c$ into the formula:
$s = \dfrac{{9 + 12 + 15}}{2}$
$s = \dfrac{{36}}{2} = 18$
Now, use the Heron’s formula to find the area of the triangle, which is given as:
Area of triangle $ADC$$ = \sqrt {s(s - a)(s - b)(s - c)} $
Substitute the values of $s,a,b,$ and $c$into the formula:
Area of triangle $ADC$$ = \sqrt {18(18 - 9)(18 - 12)(18 - 15)} $
Area of triangle $ADC$$ = \sqrt {18 \times 9 \times 6 \times 3} $
Area of triangle $ADC$$ = \sqrt {2916} $
Area of triangle $ADC$$ = 54{\text{ c}}{{\text{m}}^2}$
The area of the triangle $ADC$ is $54{\text{ c}}{{\text{m}}^2}$.
So, the area of the quadrilateral is given as:
∴ The area of quadrilateral$ = {\text{Area of triangle ABC + Area of triangle ADC}}$
The area of quadrilateral $ = \left( {10\sqrt {17} + 54} \right){\text{ c}}{{\text{m}}^2}$
Therefore, the area of the quadrilateral is $\left( {10\sqrt {17} + 54} \right){\text{ c}}{{\text{m}}^2}$.

Note:When we have given the three sides of the triangle then we can use the Heron’s formula to find the area of the triangle and when we have given the base and the height of the triangle then we can use the formula for finding the area:
Area of triangle $ = \dfrac{1}{2}\left( {Base} \right)\left( {Height} \right)$