10.4 Areas of Quadrilaterals & Complex Shapes

Since $\angle C = 90^\circ$, $\Delta BCD$ is a right-angled triangle.

Calculate $BD$ (diagonal) using Pythagoras: $BD = \sqrt{12^2 + 5^2} = \sqrt{144+25} = \sqrt{169} = 13$ m.

Area($\Delta BCD$) = $\frac{1}{2} \times 12 \times 5 = 30 \text{ m}^2$.

Now consider $\Delta ABD$. Sides are $9, 8, 13$. Use Heron's.

$s = (9+8+13)/2 = 30/2 = 15$.

Area($\Delta ABD$) = $\sqrt{15(15-9)(15-8)(15-13)} = \sqrt{15 \cdot 6 \cdot 7 \cdot 2}$.

$A = \sqrt{1260} = \sqrt{36 \times 35} = 6\sqrt{35} \approx 35.5 \text{ m}^2$.

Total Area = $30 + 35.5 = 65.5 \text{ m}^2$.