The problem requires finding the area of a quadrilateral ABCD, which is twice the area of triangle PQR. We are given the side ratios and perimeter of triangle PQR.

Determine Triangle PQR Sides

Let the sides of triangle PQR be $4x$, $5x$, and $6x$, based on the given ratio.

• Perimeter = Sum of sides
• $90 \text{ cm} = 4x + 5x + 6x$
• $90 = 15x$
• $x = \frac{90}{15} = 6$

The sides of triangle PQR are:

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

Find Triangle PQR Area using Heron's Formula

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

• $s = \frac{\text{Perimeter}}{2} = \frac{90}{2} = 45 \text{ cm}$

Now, apply Heron's formula: Area $= \sqrt{s(s-a)(s-b)(s-c)}$

• Area $= \sqrt{45(45-24)(45-30)(45-36)}$
• Area $= \sqrt{45 \times 21 \times 15 \times 9}$
• Area $= \sqrt{(9 \times 5) \times (3 \times 7) \times (3 \times 5) \times 9}$
• Area $= \sqrt{9^2 \times 5^2 \times 3^2 \times 7}$
• Area $= 9 \times 5 \times 3 \times \sqrt{7}$
• Area $= 135\sqrt{7} \text{ cm}^2$

Using the given approximation $\sqrt{7} \approx 2.6$:

• Area $\approx 135 \times 2.6$
• Area $\approx 351 \text{ cm}^2$

Determine Quadrilateral ABCD Area

The area of quadrilateral ABCD is twice the area of triangle PQR.

• Area of ABCD $= 2 \times (\text{Area of PQR})$
• Area of ABCD $\approx 2 \times 351 \text{ cm}^2$
• Area of ABCD $\approx 702 \text{ cm}^2$

Therefore, the area of quadrilateral ABCD is approximately $702 \text{ cm}^2$.