You're on a roll — Here are the solutions to the questions: Question 4: Calculate the surface area of the hexagonal prism. The base is a regular hexagon of sides $10 \text{ cm}$, and the height is $30 \text{ cm}$. Step 1: Calculate the area of one equilateral triangle forming the base using Heron's formula. A regular hexagon can be divided into 6 equilateral triangles. For an equilateral triangle with side $s = 10 \text{ cm}$: The semi-perimeter $p$ is: $$ p = \frac{10 \text{ cm} + 10 \text{ cm} + 10 \text{ cm}}{2} = \frac{30 \text{ cm}}{2} = 15 \text{ cm} $$ Using Heron's formula, the area of one equilateral triangle ($A_{triangle}$) is: $$ A_{triangle} = \sqrt{p(p-s)(p-s)(p-s)} $$ $$ A_{triangle} = \sqrt{15(15-10)(15-10)(15-10)} $$ $$ A_{triangle} = \sqrt{15 \times 5 \times 5 \times 5} $$ $$ A_{triangle} = \sqrt{15 \times 125} $$ $$ A_{triangle} = \sqrt{1875} $$ $$ A_{triangle} = \sqrt{625 \times 3} = 25\sqrt{3} \text{ cm}^2 $$ Step 2: Calculate the area of the hexagonal base ($A_{base}$). Since the hexagon consists of 6 such equilateral triangles: $$ A_{base} = 6 \times A_{triangle} $$ $$ A_{base} = 6 \times 25\sqrt{3} \text{ cm}^2 = 150\sqrt{3} \text{ cm}^2 $$ Step 3: Calculate the lateral surface area ($A_{lateral}$). The lateral surface consists of 6 rectangular faces. Each face has a width equal to the side of the hexagon ($10 \text{ cm}$) and a height equal to the prism's height ($30 \text{ cm}$). Area of one rectangular face = $10 \text{ cm} \times 30 \text{ cm} = 300 \text{ cm}^2$. $$ A_{lateral} = 6 \times 300 \text{ cm}^2 = 1800 \text{ cm}^2 $$ Step 4: Calculate the total surface area ($SA_{total}$). The total surface area is the sum of the areas of the two bases and the lateral surface area. $$ SA_{total} = 2 \times A_{base} + A_{lateral} $$ $$ SA_{total} = 2 \times 150\sqrt{3} \text{ cm}^2 + 1800 \text{ cm}^2 $$ $$ SA_{total} = 300\sqrt{3} \text{ cm}^2 + 1800 \text{ cm}^2 $$ Using $\sqrt{3} \approx 1.73205$: $$ SA_{total} \approx 300 \times 1.73205 \text{ cm}^2 + 1800 \text{ cm}^2 $$ $$ SA_{total} \approx 519.615 \text{ cm}^2 + 1800 \text{ cm}^2 $$ $$ SA_{total} \approx 2319.615 \text{ cm}^2 $$ Rounding to two decimal places: $$ SA_{total} \approx 2319.62 \text{ cm}^2 $$ $$\boxed{2319.62 \text{ cm}^2}$$ Question 5: Calculate the volume of the cone to the nearest millimetre. The cone has a radius of $5 \text{ cm}$ and a perpendicular height of $8 \text{ cm}$. Use $\pi = 3.14$. Step 1: Identify the given values. Radius ($r$) = $5 \text{ cm}$ Height ($h$) = $8 \text{ cm}$ $\pi = 3.14$ Step 2: Use the formula for the volume of a cone. The volume ($V$) of a cone is given by: $$ V = \frac{1}{3}\pi r^2 h $$ Step 3: Substitute the values and calculate the volume in cubic centimetres. $$ V = \frac{1}{3} \times 3.14 \times (5 \text{ cm})^2 \times 8 \text{ cm} $$ $$ V = \frac{1}{3} \times 3.14 \times 25 \text{ cm}^2 \times 8 \text{ cm} $$ $$ V = \frac{1}{3} \times 3.14 \times 200 \text{ cm}^3 $$ $$ V = \frac{628}{3} \text{ cm}^3 $$ $$ V \approx 209.3333... \text{ cm}^3 $$ Step 4: Convert the volume to cubic millimetres and round to the nearest whole number. Since $1 \text{ cm} = 10 \text{ mm}$, then $1 \text{ cm}^3 = (10 \text{ mm})^3 = 1000 \text{ mm}^3$. $$ V \approx 209.3333... \times 1000 \text{ mm}^3 $$ $$ V \approx 209333.333... \text{ mm}^3 $$ Rounding to the nearest whole number (nearest cubic millimetre): $$ V \approx 209333 \text{ mm}^3 $$ $$\boxed{209333 \text{ mm}^3}$$ Send me the next one 📸