1. **State the problem:** We need to find the area of quadrilateral EFGH with given side lengths: EH = 5 cm (vertical), GH = 10 cm (horizontal), FG = 2 cm (vertical), and EF is diagonal. 2. **Analyze the shape:** The quadrilateral has right angles at H and G, so EH and GH form a right angle, and FG is vertical with a right angle at G. The diagonal EG divides the quadrilateral into two triangles: EGH and EFG. 3. **Calculate the length of diagonal EG:** Since EH = 5 cm (vertical) and GH = 10 cm (horizontal), triangle EGH is right-angled at H. Use Pythagoras theorem: $$EG = \sqrt{EH^2 + GH^2} = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} = 11.2\text{ cm (to 1 d.p.)}$$ 4. **Calculate area of triangle EGH:** $$\text{Area}_{EGH} = \frac{1}{2} \times EH \times GH = \frac{1}{2} \times 5 \times 10 = 25\text{ cm}^2$$ 5. **Calculate area of triangle EFG:** FG = 2 cm (vertical), and EG = 11.2 cm (hypotenuse). We need the base EF or height to find the area. Since EF is diagonal and not given, use the fact that triangle EFG is right angled at G with FG vertical and GH horizontal. The base FG = 2 cm and the height is the horizontal distance from F to G, which is the same as GH = 10 cm. Area of triangle EFG: $$\text{Area}_{EFG} = \frac{1}{2} \times FG \times GH = \frac{1}{2} \times 2 \times 10 = 10\text{ cm}^2$$ 6. **Total area of quadrilateral EFGH:** $$\text{Area} = \text{Area}_{EGH} + \text{Area}_{EFG} = 25 + 10 = 35\text{ cm}^2$$ **Final answer:** The area of quadrilateral EFGH is **35.0 cm²**.