1. **Problem 7:** Find the area of figure ABCDEFGH. Given: - AC = HF = $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ cm. - Area = area of rectangle ADEH + 2 × area of triangle ABC. Step 1: Calculate area of rectangle ADEH. - Rectangle ADEH has length DE = 8 cm (given) and height AH = 5 cm. - Area = length × height = $8 \times 5 = 40$ cm$^2$. Step 2: Calculate area of triangle ABC. - Triangle ABC is right-angled with base BC = 4 cm and height AB = 5 cm. - Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10$ cm$^2$. Step 3: Calculate total area. - Total area = area of rectangle + 2 × area of triangle - $= 40 + 2 \times 10 = 40 + 20 = 60$ cm$^2$. 2. **Problem 8:** Find the area of a regular hexagon ABCDEF with each side 13 cm and height 23 cm. Given: - Height AL + LM + DM = 23 cm, - LM = BC = 13 cm (side length), - AL = DM = x. - So: $x + 13 + x = 23 \Rightarrow 2x = 10 \Rightarrow x = 5$ cm. Step 1: Find length FL. - Using Pythagoras theorem for triangle AFL: $$FL^2 = AF^2 - AL^2 = 13^2 - 5^2 = 169 - 25 = 144$$ - So, $FL = \sqrt{144} = 12$ cm. Step 2: Area of trapezium ADEF. - Bases AF and DE are both equal to 13 cm (side of hexagon), height FL = 12 cm. - Area of trapezium $= \frac{1}{2} \times (AF + DE) \times FL = \frac{1}{2} \times (13 + 13) \times 12 = \frac{1}{2} \times 26 \times 12 = 156$ cm$^2$. Step 3: Total area of hexagon. - Hexagon = 2 × area of trapezium ADEF - Total area $= 2 \times 156 = 312$ cm$^2$. **Final answers:** - Question 7 area = $60$ cm$^2$. - Question 8 area = $312$ cm$^2$.