Kite Area 83486A

1. **Problem 1: Area of the kite** We are given a kite-shaped quadrilateral with diagonals measuring $25$ cm and $40$ cm, and a vertical height of $60$ cm (likely the length of one diagonal). The area of a kite can be calculated using the formula: $$\text{Area} = \frac{1}{2} \times d_1 \times d_2$$ where $d_1$ and $d_2$ are the lengths of the diagonals. 2. **Calculate the area of the kite:** $$\text{Area} = \frac{1}{2} \times 25 \times 40$$ $$= \frac{1}{2} \times 1000$$ $$= 500 \text{ cm}^2$$ So, the area of the kite is $500$ square centimeters. --- 3. **Problem 2: Area of triangle ABC (national park)** Given: - Side $AB = 21.5$ km - Side $AC = 14.2$ km - Bearing of $B$ from $A$ is $324^\circ$ - Bearing of $C$ from $A$ is $42^\circ$ 4. **Find the angle $BAC$ between sides $AB$ and $AC$:** The angle between the two bearings is: $$\theta = 42^\circ - 324^\circ + 360^\circ = 78^\circ$$ (We add $360^\circ$ to get a positive angle difference.) 5. **Use the formula for the area of a triangle given two sides and the included angle:** $$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\theta)$$ 6. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times 21.5 \times 14.2 \times \sin(78^\circ)$$ Calculate $\sin(78^\circ) \approx 0.9781$: $$\text{Area} = \frac{1}{2} \times 21.5 \times 14.2 \times 0.9781$$ $$= 0.5 \times 21.5 \times 14.2 \times 0.9781$$ $$= 149.2 \text{ km}^2$$ **Final answers:** - Area of the kite: $500$ cm$^2$ - Area of the national park (triangle ABC): approximately $149.2$ km$^2$