Find the area of a quadrilateral ABCD whose area is twice the area of a triangle PQR. The sides of triangle PQR are in the ratio 4:5:6 and the perimeter of the triangle is 90 cm (use ).

The correct option is 702 cm².

Step 1: Understand the given data
We are given that the sides of the triangle PQR are in the ratio $4:5:6$.
The perimeter of triangle PQR is $90\text{cm}$.
The area of quadrilateral ABCD is twice the area of triangle PQR.

Step 2: Find the side lengths of triangle PQR
Let the side lengths of triangle PQR be $a=4x$, $b=5x$, and $c=6x$, where $x$ is a constant ratio multiplier.

Since the perimeter of triangle PQR is equal to the sum of its three sides:

$a+b+c=90$

$4x+5x+6x=90$

$15x=90$

$x=\frac{90}{15}=6$

Now, substitute $x=6$ back into the side expressions to find the actual lengths of the sides:

$a=4\times 6=24\text{cm}$
$b=5\times 6=30\text{cm}$
$c=6\times 6=36\text{cm}$

Step 3: Calculate the area of triangle PQR using Heron's Formula
First, find the semi-perimeter ($s$) of triangle PQR:

$s=\frac{a+b+c}{2}=\frac{90}{2}=45\text{cm}$

According to Heron's Formula, the area of a triangle is given by:

$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$

Substitute the values of $s$, $a$, $b$, and $c$ into the formula:

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{45(45-24)(45-30)(45-36)}$

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{45\times 21\times 15\times 9}$

Factorize the numbers inside the square root to simplify:

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{(15\times 3)\times (3\times 7)\times 15\times 9}$

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{{15}^2\times 3^2\times 3^2\times 7}$

$\text{Area of}\Delta \mathrm{PQR}=15\times 3\times 3\times \sqrt{7}$

$\text{Area of}\Delta \mathrm{PQR}=135\sqrt{7}{\text{cm}}^2$

Approximating $\sqrt{7}\approx 2.6$ (or $2.60$):

$\text{Area of}\Delta \mathrm{PQR}\approx 135\times 2.6=351{\text{cm}}^2$

Step 4: Find the area of quadrilateral ABCD
We are given that the area of quadrilateral ABCD is twice the area of triangle PQR:

$\text{Area of ABCD}=2\times \text{Area of}\Delta \mathrm{PQR}$

$\text{Area of ABCD}=2\times 351=702{\text{cm}}^2$

Therefore, the area of quadrilateral ABCD is 702 cm².