1. **Problem statement:** Calculate the area of trapezium ABCD, the perpendicular distance $h$ from point B to line AD, and angle DAB given: - $AB=9$ cm, $BC=4$ cm, $CD=6$ cm, $DA=5$ cm - Angles $ABC$ and $BCD$ are right angles 2. **Step 1: Find the area of trapezium ABCD** The area of a trapezium is given by: $$\text{Area} = \frac{(\text{sum of parallel sides}) \times \text{height}}{2}$$ Here, $AB$ and $CD$ are parallel (both horizontal), and the height is $BC=4$ cm. So, $$\text{Area} = \frac{(9 + 6) \times 4}{2} = \frac{15 \times 4}{2} = 30 \text{ cm}^2$$ 3. **Step 2: Find the perpendicular distance $h$ from B to AD** We use coordinate geometry for clarity: - Place $A$ at origin $(0,0)$ - Since $AB=9$ cm horizontal, $B$ is at $(9,0)$ - $BC=4$ vertical, so $C$ is at $(9,4)$ - $CD=6$ horizontal, so $D$ is at $(3,4)$ (since $C$ is at 9,4 and $CD=6$ to the left) Line $AD$ passes through $A(0,0)$ and $D(3,4)$. Equation of $AD$: Slope $m = \frac{4-0}{3-0} = \frac{4}{3}$ Equation: $$y = \frac{4}{3}x$$ Distance from point $B(9,0)$ to line $AD$ is: $$h = \frac{|4/3 \times 9 - 0 + 0|}{\sqrt{(4/3)^2 + 1^2}} = \frac{12}{\sqrt{\frac{16}{9} + 1}} = \frac{12}{\sqrt{\frac{25}{9}}} = \frac{12}{\frac{5}{3}} = 7.2 \text{ cm}$$ 4. **Step 3: Find angle DAB** Angle $DAB$ is the angle between vectors $AB$ and $AD$ at point $A$. Vectors: $$\vec{AB} = (9,0)$$ $$\vec{AD} = (3,4)$$ Use dot product formula: $$\vec{AB} \cdot \vec{AD} = |\vec{AB}| |\vec{AD}| \cos \theta$$ Calculate magnitudes: $$|\vec{AB}| = 9$$ $$|\vec{AD}| = \sqrt{3^2 + 4^2} = 5$$ Dot product: $$9 \times 3 + 0 \times 4 = 27$$ So, $$\cos \theta = \frac{27}{9 \times 5} = \frac{27}{45} = 0.6$$ Therefore, $$\theta = \cos^{-1}(0.6) \approx 53.13^\circ$$ **Final answers:** - (i) Area = 30 cm$^2$ - (ii) Perpendicular distance $h = 7.2$ cm - (iii) Angle $DAB \approx 53.13^\circ$