1. **Problem statement:** We have a trapezium ABCD with parallel sides AB and CD. Given: AB = 12 cm, CD = 8 cm, height = 6 cm. Diagonal AC divides the trapezium into triangles △ABC and △ADC. We need to find: a) Area of trapezium ABCD. b) Perimeter of △ABC given BC = 10 cm. c) Perimeter of △ADC given AD = 9 cm. 2. **Formula for area of trapezium:** $$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$ 3. **Calculate area:** $$\text{Area} = \frac{1}{2} \times (12 + 8) \times 6 = \frac{1}{2} \times 20 \times 6 = 10 \times 6 = 60 \text{ cm}^2$$ 4. **Calculate length of diagonal AC:** Since ABCD is a trapezium with height 6 cm, and AB and CD are parallel, we can place it on coordinate axes for clarity: - Let D be at (0,0), C at (8,0) (since CD=8). - Height is 6, so A and B lie on line y=6. - AB=12, so if A is at (x,6), B is at (x+12,6). - Since ABCD is a trapezium, points A and D are connected by side AD, and B and C by BC. We know BC = 10 cm. Coordinates: - C = (8,0) - B = (x+12,6) Distance BC: $$BC = \sqrt{(x+12 - 8)^2 + (6 - 0)^2} = 10$$ Simplify: $$\sqrt{(x+4)^2 + 36} = 10$$ Square both sides: $$(x+4)^2 + 36 = 100$$ $$(x+4)^2 = 64$$ $$x+4 = \pm 8$$ Two solutions: - $x = 4$ or $x = -12$ Choose $x=4$ (so A at (4,6), B at (16,6)) to keep trapezium shape. 5. **Calculate length of diagonal AC:** Coordinates: - A = (4,6) - C = (8,0) $$AC = \sqrt{(8-4)^2 + (0-6)^2} = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \approx 7.21 \text{ cm}$$ 6. **Calculate perimeter of △ABC:** Sides: AB = 12 cm, BC = 10 cm, AC = $2\sqrt{13}$ cm. $$\text{Perimeter} = AB + BC + AC = 12 + 10 + 2\sqrt{13} \approx 12 + 10 + 7.21 = 29.21 \text{ cm}$$ 7. **Calculate perimeter of △ADC:** Given AD = 9 cm. Sides: AD = 9 cm, DC = 8 cm, AC = $2\sqrt{13}$ cm. $$\text{Perimeter} = AD + DC + AC = 9 + 8 + 2\sqrt{13} \approx 9 + 8 + 7.21 = 24.21 \text{ cm}$$