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The diagram shows a frustum of a cone with base diameter and top diameter of 24cm and 6cm respectively. The height of the frustum is 15cm. [Take π as 3.142]. Calculate the (i) height of the cone, (ii) volume of the frustum.

Mathematics

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Answers

2 parts
a(i)

20 cm20 \text{ cm}

a(ii)

2969.19 cm32969.19 \text{ cm}^3

Step-by-step solution

6 steps

Here's how to solve the problem:

Part a(i)· 3 steps

height of the cone

  1. Identify the radii and frustum height.

    The base diameter is 24 cm24 \text{ cm}, so the base radius is R=242=12 cmR = \frac{24}{2} = 12 \text{ cm}.
    The top diameter is 6 cm6 \text{ cm}, so the top radius is r=62=3 cmr = \frac{6}{2} = 3 \text{ cm}.
    The height of the frustum is Hf=15 cmH_f = 15 \text{ cm}.
    Let HH be the height of the original large cone and hh be the height of the small cone that was removed.
    We know that Hf=H−hH_f = H - h, so 15=H−h15 = H - h, which means H=h+15H = h + 15.

  2. Use similar triangles to find the height of the small cone (hh).

    Consider the cross-section of the cone. The ratio of the radii to their respective heights in similar triangles is constant:

    RH=rh\frac{R}{H} = \frac{r}{h}

    Substitute the known values:

    12H=3h\frac{12}{H} = \frac{3}{h}

    Now substitute H=h+15H = h + 15:

    12h+15=3h\frac{12}{h + 15} = \frac{3}{h}

    Cross-multiply to solve for hh:

    12h=3(h+15)12h = 3(h + 15) 12h=3h+4512h = 3h + 45 12h−3h=4512h - 3h = 45 9h=459h = 45 h=459=5 cmh = \frac{45}{9} = 5 \text{ cm}

  3. Calculate the height of the original large cone (HH).

    H=h+15H = h + 15 H=5+15H = 5 + 15 H=20 cmH = 20 \text{ cm}

    The height of the cone is 20 cm\boxed{20 \text{ cm}}.

Part a(ii)

2969.19 cm32969.19 \text{ cm}^3

volume of the frustum

  1. Calculate the volume of the large cone (VLV_L).

    The radius of the large cone is R=12 cmR = 12 \text{ cm} and its height is H=20 cmH = 20 \text{ cm}. Use π=3.142\pi = 3.142.
    The formula for the volume of a cone is V=13πr2hV = \frac{1}{3} \pi r^2 h.

    VL=13πR2HV_L = \frac{1}{3} \pi R^2 H VL=13×3.142×(12)2×20V_L = \frac{1}{3} \times 3.142 \times (12)^2 \times 20 VL=13×3.142×144×20V_L = \frac{1}{3} \times 3.142 \times 144 \times 20 VL=3.142×48×20V_L = 3.142 \times 48 \times 20 VL=3.142×960V_L = 3.142 \times 960 VL=3016.32 cm3V_L = 3016.32 \text{ cm}^3

  2. Calculate the volume of the small cone (VSV_S).

    The radius of the small cone is r=3 cmr = 3 \text{ cm} and its height is h=5 cmh = 5 \text{ cm}.

    VS=13πr2hV_S = \frac{1}{3} \pi r^2 h VS=13×3.142×(3)2×5V_S = \frac{1}{3} \times 3.142 \times (3)^2 \times 5 VS=13×3.142×9×5V_S = \frac{1}{3} \times 3.142 \times 9 \times 5 VS=3.142×3×5V_S = 3.142 \times 3 \times 5 VS=3.142×15V_S = 3.142 \times 15 VS=47.13 cm3V_S = 47.13 \text{ cm}^3

  3. Calculate the volume of the frustum (VfV_f).

    The volume of the frustum is the volume of the large cone minus the volume of the small cone.

    Vf=VL−VSV_f = V_L - V_S Vf=3016.32−47.13V_f = 3016.32 - 47.13 Vf=2969.19 cm3V_f = 2969.19 \text{ cm}^3

    The volume of the frustum is 2969.19 cm3\boxed{2969.19 \text{ cm}^3}.

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