Question

If the vertices of the quadrilateral is given by ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$ then area of the quadrilateral is:
A. $36$
B. $24$
C. $16$
D. $81$

Answer 1

Here we are given an equation representing the vertices of the quadrilateral that is ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$ and we are asked to find the area of the quadrilateral.so while approaching such kind of questions we first need to take a look at the given equation in the question as it is the only main source through which the vertices of the quadrilateral can be found out like in this question the equation is provided to us is ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$ so the vertices can be calculated easily by equating each of the terms containing $x$ and $y$ to zero and simplifying the terms to get the vertices and plotting them and finding the area of the resultant quadrilateral.

Complete step by step solution:
Here we are given an equation signifying the vertices of the quadrilateral that is ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$ and we are asked to find the area of the quadrilateral.
For the area we need to first find out the vertices of the quadrilateral which can be done as in the given equation ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$ we can see that the sum of the two squares of equation is equal to zero
So each of the terms of squares that is ${\left( {{x^2} - 4} \right)^2}and{\left( {{y^2} - 9} \right)^2}$ are positive and are equal to zero therefore both have to be equal to zero respectively so by equating ${\left( {{x^2} - 4} \right)^2}and{\left( {{y^2} - 9} \right)^2}$ equal to zero we get –

$${\left( {{x^2} - 4} \right)^2} = 0 \\\ x = \pm 2 \\\$$

And

$${\left( {{y^2} - 9} \right)^2} = 0 \\\ y = \pm 3 \\\$$

Now plotting these values of $xandy$that is $\left( {2,3} \right),\left( {2, - 3} \right),\left( { - 2, - 3} \right),\left( { - 2,3} \right)$we would get –

Here it is clearly seen that by plotting the points the resulting quadrilateral represented by the equation ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$is rectangle with length can be calculated by the distance formula that is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
Here $\left( {{x_1},{y_1}} \right) =$coordinates of the first point
And $\left( {{x_2},{y_2}} \right) =$coordinates of the second point
Now calculating the distance between the points $\left( {2,3} \right),\left( { - 2,3} \right)$for the length of the triangle by applying the distance formula we get-
$\sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {3 - 3} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 0 \right)}^2}}$
$\Rightarrow \sqrt {16 + {{\left( 0 \right)}^2}}$
$= 4$which is the length of the rectangle
Similarly for the breath of the rectangle we need to calculate the distance between the points $\left( {2, - 3} \right),\left( { - 2, - 3} \right)$using the distance formula which is describe above –

$$\sqrt {{{\left( {2 - 2} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} \\\ \Rightarrow \sqrt {{{\left( 0 \right)}^2} + {{\left( { - 6} \right)}^2}} \\\ \Rightarrow \sqrt {36} \\\ = 6 \\\$$

So, it is the breadth of the rectangle.
Now for the area of the rectangle formed by the equation ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$ we would use the formula for the area of the rectangle that is $\text{length} \times \text{breadth}$
So the area of the rectangle is $4 \times 6 = 24s{q^2}\text{unit}$
So the area of the quadrilateral of the equation ${\left( {{x^2} - 4} \right)^2} + {\left( {{y^2} - 9} \right)^2} = 0$is $24s{q^2}unit$ which is same as option B that is $24$.

Hence the correct option is B.

Note:
While solving such kind of questions we need to keep In mind to find the vertices of the quadrilateral which would make the job of finding the area very easy as we can plot them and find which kind of quadrilateral is it and apply the formula of the area associated to that type of the quadrilateral and one another thing the person should know that is the distance formula which is the driving concept of this question.