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.

The correct answer is 702 cm².

Step 1: Determine the side lengths of triangle PQR

Let the sides of triangle PQR be $a$, $b$, and $c$.

Given that the ratio of the sides is $4:5:6$, we can express the sides as:

$a=4x$

$b=5x$

$c=6x$

where $x$ is a constant scaling factor.

The perimeter of triangle PQR is given as 90 cm. Summing the sides:

$a+b+c=90$

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

$15x=90$

$x=6$

Now, substituting $x=6$ back 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 2: Calculate the area of triangle PQR using Heron's Formula

First, find the semi-perimeter ($s$) of triangle PQR:

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

Using Heron's formula for the area of a triangle:

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{s(s-a)(s-b)(s-c)}$

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

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

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

Breaking down into prime factors:

$45=9\times 5$

$21=7\times 3$

$15=5\times 3$

Substituting back under the square root:

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{(9\times 5)\times (7\times 3)\times (5\times 3)\times 9}$

$\text{Area of}\Delta \mathrm{PQR}=\sqrt{9^2\times 5^2\times 3^2\times 7}$

$\text{Area of}\Delta \mathrm{PQR}=9\times 5\times 3\times \sqrt{7}=135\sqrt{7}{\text{cm}}^2$

Using the approximate value $\sqrt{7}\approx 2.6$:

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

Step 3: Calculate the area of quadrilateral ABCD

According to the problem, the area of quadrilateral ABCD is twice the area of triangle PQR:

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

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