1. **Find the area of the figure ABCDEFGH.** Given: Rectangle ADEH with dimensions AD = 8 cm and DE = 4 cm. Triangles ABC and FGH with bases 4 cm and heights 5 cm each. Step 1: Calculate the area of rectangle ADEH. $$ \text{Area}_{rectangle} = AD \times DE = 8 \times 4 = 32 \text{ cm}^2 $$ Step 2: Calculate the area of one triangle, ABC. Base = 4 cm, Height = 5 cm $$ \text{Area}_{triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10 \text{ cm}^2 $$ Step 3: There are two such triangles (ABC and FGH), so total area of triangles is: $$ 2 \times 10 = 20 \text{ cm}^2 $$ Step 4: Total area of figure ABCDEFGH is area of rectangle plus area of two triangles: $$ 32 + 20 = 52 \text{ cm}^2 $$ --- 2. **Find the area of the regular hexagon ABCDEF with side 13 cm and height 23 cm.** Given: Each side = 13 cm, height (distance between parallel sides) = 23 cm. Step 1: Let AL = DM = x, LM = BC = 13 cm (side length), and height AL + BC + DM = 23 cm. So, $$ x + 13 + x = 23 \implies 2x = 10 \implies x = 5 \text{ cm} $$ Step 2: Find FL (the height of trapezium ADEF) using Pythagoras theorem: $$ FL = \sqrt{AF^2 - AL^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm} $$ Step 3: Area of trapezium ADEF: Parallel sides: AD = EF = 23 cm (height), and lengths AL + DM = 5 + 5 = 10 cm (the other base), but in trapezium, the parallel sides are FL and DE. Actually, trapezium ADEF has bases AL + DM + BC = 23 cm (height) and side = 13 cm. But to keep simple, area is given by: $$ \text{Area}_{trap} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (13 + 23) \times 12 = \frac{1}{2} \times 36 \times 12 = 216 \text{ cm}^2 $$ Step 4: Area of hexagon = 2 × area of trapezium ADEF = $$ 2 \times 216 = 432 \text{ cm}^2 $$ **Final answers:** - Area of figure ABCDEFGH = $52$ cm$^{2}$ - Area of regular hexagon ABCDEF = $432$ cm$^{2}$