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In the diagram ABCDEF is a regular hexagon, the length of each side is 5 cm. It is the base of a hexagonal prism whose height is 12cm. Calculate the length of BD, area of ABCDEF, volume of the prism.

Mathematics

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Answers

3 parts
i

8.66 cm (to 3 s.f.)8.66 \text{ cm (to 3 s.f.)}

ii

65.0 cm2 (to 3 s.f.)65.0 \text{ cm}^2 \text{ (to 3 s.f.)}

iii

779 cm3 (to 3 s.f.)779 \text{ cm}^3 \text{ (to 3 s.f.)}

Step-by-step solution

9 steps

Here's the solution for the questions about the regular hexagon and prism:

Given:
The diagram ABCDEF is a regular hexagon.
Length of each side, s=5s = 5 cm.
Height of the hexagonal prism, h=12h = 12 cm.

Part i· 3 steps

Calculate the length of BD

  1. Identify the properties of a regular hexagon.

    A regular hexagon has interior angles of 120∘120^\circ. Consider triangle BCD.
    Sides BC and CD are sides of the hexagon, so BC=CD=5BC = CD = 5 cm.
    The angle ∠BCD=120∘\angle BCD = 120^\circ.

  2. Apply the Law of Cosines to triangle BCD.

    The Law of Cosines states a2=b2+c2−2bccos⁡(A)a^2 = b^2 + c^2 - 2bc \cos(A).
    Here, BD2=BC2+CD2−2(BC)(CD)cos⁡(∠BCD)BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD).

    BD2=52+52−2(5)(5)cos⁡(120∘)BD^2 = 5^2 + 5^2 - 2(5)(5)\cos(120^\circ) BD2=25+25−50(−12)BD^2 = 25 + 25 - 50\left(-\frac{1}{2}\right) BD2=50+25BD^2 = 50 + 25 BD2=75BD^2 = 75

  3. Solve for BD.

    BD=75=25×3=53 cmBD = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \text{ cm} BD≈5×1.73205BD \approx 5 \times 1.73205 BD≈8.66 cm (to 3 s.f.)BD \approx \boxed{8.66 \text{ cm (to 3 s.f.)}}

Part ii

65.0 cm2 (to 3 s.f.)65.0 \text{ cm}^2 \text{ (to 3 s.f.)}

Calculate the area of ABCDEF

  1. Recall the formula for the area of a regular hexagon.

    A regular hexagon can be divided into 6 equilateral triangles, each with side length ss.
    The area of one equilateral triangle is 34s2\frac{\sqrt{3}}{4}s^2.
    The area of the hexagon is 6×34s26 \times \frac{\sqrt{3}}{4}s^2.

  2. Substitute the side length s=5s = 5 cm into the formula.

    Area=6×34(52)\text{Area} = 6 \times \frac{\sqrt{3}}{4} (5^2) Area=6×34×25\text{Area} = 6 \times \frac{\sqrt{3}}{4} \times 25 Area=15034\text{Area} = \frac{150\sqrt{3}}{4} Area=7532 cm2\text{Area} = \frac{75\sqrt{3}}{2} \text{ cm}^2

  3. Calculate the numerical value and round to 3 significant figures.

    Area≈75×1.732052\text{Area} \approx \frac{75 \times 1.73205}{2} Area≈129.903752\text{Area} \approx \frac{129.90375}{2} Area≈65.0 cm2 (to 3 s.f.)\text{Area} \approx \boxed{65.0 \text{ cm}^2 \text{ (to 3 s.f.)}}

Part iii

779 cm3 (to 3 s.f.)779 \text{ cm}^3 \text{ (to 3 s.f.)}

Calculate the volume of the prism

  1. Recall the formula for the volume of a prism.

    Volume of a prism = Base Area ×\times Height.

  2. Substitute the calculated base area and given height.

    Base Area = Area of ABCDEF = 7532\frac{75\sqrt{3}}{2} cm2^2.
    Height, h=12h = 12 cm.

    Volume=(7532)×12\text{Volume} = \left(\frac{75\sqrt{3}}{2}\right) \times 12 Volume=753×6\text{Volume} = 75\sqrt{3} \times 6 Volume=4503 cm3\text{Volume} = 450\sqrt{3} \text{ cm}^3

  3. Calculate the numerical value and round to 3 significant figures.

    Volume≈450×1.73205\text{Volume} \approx 450 \times 1.73205 Volume≈779.4225\text{Volume} \approx 779.4225 Volume≈779 cm3 (to 3 s.f.)\text{Volume} \approx \boxed{779 \text{ cm}^3 \text{ (to 3 s.f.)}}
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