A model consists of a solid hemisphere attached to a solid cylinder as shown in diagram. The height of the cylinder is 26 cm and the base area is 198 cm^2.

  1. (a)

    Calculate, giving your answers to the nearest whole number, the volume of

    1. (i)

      the cylinder.

      Answer ...

    2. (ii)

      the hemisphere.

      Answer ...

  2. (b)

    Part of the cylinder in the shape of a right circular cone is removed as shown in the diagram 2.

    1. (i)

      Given that the volume of the cone removed from the model is 630 cm^3, calculate the height of the cone.

      Answer ... cm

    2. (ii)

      Calculate the total surface area of the model as shown in diagram 2, giving the answer to the nearest whole number.

      Answer ... cm^2

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Worked solution

8(a)(i)

1. Calculate the volume of the cylinder using its base area and height.

Volume of cylinder=base area×height\text{Volume of cylinder}=\text{base area}\times\text{height}
=198×26=5148 cm3=198\times 26=5148\text{ cm}^3

Answer

5148 cm35148\text{ cm}^3

8(a)(ii)

1. Find the radius of the hemisphere using the base area of the cylinder.

Since the hemisphere has the same radius as the cylinder base,
πr2=198\pi r^2=198
r2=198πr^2=\frac{198}{\pi}
r=198πr=\sqrt{\frac{198}{\pi}}

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

Volume of hemisphere
=23πr3=\frac{2}{3}\pi r^3
=23π(198π)3=\frac{2}{3}\pi\left(\sqrt{\frac{198}{\pi}}\right)^3

3. Calculate and round the volume to the nearest whole number.

23π(198π)3=1056.9…\frac{2}{3}\pi\left(\sqrt{\frac{198}{\pi}}\right)^3=1056.9\ldots
To the nearest whole number,
1056.9…≈10571056.9\ldots\approx 1057

Answer

1057 cm31057\text{ cm}^3

8(b)(i)

1. Substitute the cone volume and base area into the cone volume formula, then solve for the height.

For a cone,
V=13×base area×hV=\frac{1}{3}\times \text{base area}\times h
Given V=630V=630 and base area =198=198,
630=13(198)h630=\frac{1}{3}(198)h
h=630×3198=9.545…h=\frac{630\times 3}{198}=9.545\ldots
So the height of the cone is 9.55 cm9.55\text{ cm} to 3 significant figures.

Answer

9.55 cm9.55\text{ cm}

8(b)(ii)

1. Find the common radius of the cylinder, hemisphere and cone.

Since the base area of the cylinder is 198 cm2198\text{ cm}^2,
πr2=198\pi r^2=198
r=198π=7.9377… cmr=\sqrt{\frac{198}{\pi}}=7.9377\ldots\text{ cm}

2. Find the slant height of the cone removed.

From part (b)(i), the height of the cone is
h=3(630)198=9.5454… cmh=\frac{3(630)}{198}=9.5454\ldots\text{ cm}
Using Pythagoras' theorem,
l=r2+h2l=\sqrt{r^2+h^2}
l=7.9377…2+9.5454…2=12.4153… cml=\sqrt{7.9377\ldots^2+9.5454\ldots^2}=12.4153\ldots\text{ cm}

3. Write the total surface area as the sum of all exposed surfaces.

Exposed surfaces are:

  • curved surface area of cylinder: 2πrh2\pi rh
  • curved surface area of hemisphere: 2πr22\pi r^2
  • circular base of cylinder: πr2\pi r^2
  • curved surface area of conical hole: πrl\pi rl

So
TSA=2πr(26)+2πr2+πr2+πrl.\text{TSA}=2\pi r(26)+2\pi r^2+\pi r^2+\pi rl.

4. Calculate the total surface area and round to the nearest whole number.

TSA=2π(7.9377…)(26)+2(198)+198+π(7.9377…)(12.4153…)\text{TSA}=2\pi(7.9377\ldots)(26)+2(198)+198+\pi(7.9377\ldots)(12.4153\ldots)
=1296.79…+396+198+309.65…=1296.79\ldots+396+198+309.65\ldots
=2200.44…=2200.44\ldots
To the nearest whole number,
TSA=2200 cm2.\text{TSA}=2200\text{ cm}^2.

Answer

2200 cm22200\text{ cm}^2

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