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Calculate the surface area of the hexagonal prism. The base is a regular hexagon of sides 10 cm, and the height is 30 cm.

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2 parts
4

2319.62 cm22319.62 \text{ cm}^2

5

209333 mm3209333 \text{ mm}^3

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8 steps

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Part 4· 4 steps

Calculate the surface area of the hexagonal prism. The base is a regular hexagon of sides 10 cm10 \text{ cm}, and the height is 30 cm30 \text{ cm}.

  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 cms = 10 \text{ cm}:
    The semi-perimeter pp is:

    p=10 cm+10 cm+10 cm2=30 cm2=15 cmp = \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 (AtriangleA_{triangle}) is:

    Atriangle=p(p−s)(p−s)(p−s)A_{triangle} = \sqrt{p(p-s)(p-s)(p-s)} Atriangle=15(15−10)(15−10)(15−10)A_{triangle} = \sqrt{15(15-10)(15-10)(15-10)} Atriangle=15×5×5×5A_{triangle} = \sqrt{15 \times 5 \times 5 \times 5} Atriangle=15×125A_{triangle} = \sqrt{15 \times 125} Atriangle=1875A_{triangle} = \sqrt{1875} Atriangle=625×3=253 cm2A_{triangle} = \sqrt{625 \times 3} = 25\sqrt{3} \text{ cm}^2

  2. Calculate the area of the hexagonal base (AbaseA_{base}).

    Since the hexagon consists of 6 such equilateral triangles:

    Abase=6×AtriangleA_{base} = 6 \times A_{triangle} Abase=6×253 cm2=1503 cm2A_{base} = 6 \times 25\sqrt{3} \text{ cm}^2 = 150\sqrt{3} \text{ cm}^2

  3. Calculate the lateral surface area (AlateralA_{lateral}).

    The lateral surface consists of 6 rectangular faces. Each face has a width equal to the side of the hexagon (10 cm10 \text{ cm}) and a height equal to the prism's height (30 cm30 \text{ cm}).
    Area of one rectangular face = 10 cm×30 cm=300 cm210 \text{ cm} \times 30 \text{ cm} = 300 \text{ cm}^2.

    Alateral=6×300 cm2=1800 cm2A_{lateral} = 6 \times 300 \text{ cm}^2 = 1800 \text{ cm}^2

  4. Calculate the total surface area (SAtotalSA_{total}).

    The total surface area is the sum of the areas of the two bases and the lateral surface area.

    SAtotal=2×Abase+AlateralSA_{total} = 2 \times A_{base} + A_{lateral} SAtotal=2×1503 cm2+1800 cm2SA_{total} = 2 \times 150\sqrt{3} \text{ cm}^2 + 1800 \text{ cm}^2 SAtotal=3003 cm2+1800 cm2SA_{total} = 300\sqrt{3} \text{ cm}^2 + 1800 \text{ cm}^2

    Using 3≈1.73205\sqrt{3} \approx 1.73205:

    SAtotal≈300×1.73205 cm2+1800 cm2SA_{total} \approx 300 \times 1.73205 \text{ cm}^2 + 1800 \text{ cm}^2 SAtotal≈519.615 cm2+1800 cm2SA_{total} \approx 519.615 \text{ cm}^2 + 1800 \text{ cm}^2 SAtotal≈2319.615 cm2SA_{total} \approx 2319.615 \text{ cm}^2

    Rounding to two decimal places:

    SAtotal≈2319.62 cm2SA_{total} \approx 2319.62 \text{ cm}^2 2319.62 cm2\boxed{2319.62 \text{ cm}^2}

Part 5

209333 mm3209333 \text{ mm}^3

Calculate the volume of the cone to the nearest millimetre. The cone has a radius of 5 cm5 \text{ cm} and a perpendicular height of 8 cm8 \text{ cm}. Use π=3.14\pi = 3.14.

  1. Identify the given values.

    Radius (rr) = 5 cm5 \text{ cm}
    Height (hh) = 8 cm8 \text{ cm}
    π=3.14\pi = 3.14

  2. Use the formula for the volume of a cone.

    The volume (VV) of a cone is given by:

    V=13πr2hV = \frac{1}{3}\pi r^2 h

  3. Substitute the values and calculate the volume in cubic centimetres.

    V=13×3.14×(5 cm)2×8 cmV = \frac{1}{3} \times 3.14 \times (5 \text{ cm})^2 \times 8 \text{ cm} V=13×3.14×25 cm2×8 cmV = \frac{1}{3} \times 3.14 \times 25 \text{ cm}^2 \times 8 \text{ cm} V=13×3.14×200 cm3V = \frac{1}{3} \times 3.14 \times 200 \text{ cm}^3 V=6283 cm3V = \frac{628}{3} \text{ cm}^3 V≈209.3333... cm3V \approx 209.3333... \text{ cm}^3

  4. Convert the volume to cubic millimetres and round to the nearest whole number.

    Since 1 cm=10 mm1 \text{ cm} = 10 \text{ mm}, then 1 cm3=(10 mm)3=1000 mm31 \text{ cm}^3 = (10 \text{ mm})^3 = 1000 \text{ mm}^3.

    V≈209.3333...×1000 mm3V \approx 209.3333... \times 1000 \text{ mm}^3 V≈209333.333... mm3V \approx 209333.333... \text{ mm}^3

    Rounding to the nearest whole number (nearest cubic millimetre):

    V≈209333 mm3V \approx 209333 \text{ mm}^3 209333 mm3\boxed{209333 \text{ mm}^3}
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