Volume and Surface Area: past paper questions & worked solutions

$$r = \\frac{24}{2} = 12 \\text{ cm}$$

Surface Area = 4*π**r*²  = 4 × *π* × 12²  = 4 × *π* × 144  = 576*π* ≈ 1809.557... cm²

**Step 3: Round to 3 significant figures.** 1809.557 → 1810 (the 1, 8, and 1 are the first three sig figs).

Use the stated radius, not a diameter.

Prism volume equals cross-sectional area times length; equate the two volumes.

For a cylinder, \[ V=\pi r^2h \]

Capacity of the cuboid: \[ 35\times 28\times 20=19600\text{ cm}^3 \]

Water already in the cuboid: \[ 35\times 28\times 9=8820\text{ cm}^3 \]

Volume of the full cylinder: \[ \pi(10)^2(33)=10367.255\ldots\text{ cm}^3 \]

Total water after pouring: \[ 8820+10367.255\ldots=19187.255\ldots \]

Since \[ 19187.255\ldots<19600 \]

the cuboid will not be completely full.

Its volume is \(1014\text{ cm}^3\), so

Substitute \(y=\frac{169}{x^2}\):

Area of one trapezium:

Perimeter of the trapezium:

Rectangular faces:

Total surface area:

Surface area:

The triangular end is a right triangle with sides \(6\), \(8\), and \(10\).

Area of the two triangular ends: \[ 2\left(\frac12\times 6\times 8\right)=48 \]

Area of the three rectangles: \[ 15(6+8+10)=360 \]

Total surface area: \[ 48+360=408 \]

The bowl is a hemispherical shell.

The volume of the metal is:

Volume of a hemisphere:

So

Area of the trapezium cross-section: \[ \frac12(7+12)h=9.5h \]

Volume of the prism: \[ 9.5h\times 10=608 \]

Keep \(h\) unrounded; using \(h=17\) gives \(1696\).

The volume is the product of the three perpendicular edge lengths.

There are two faces of each of the three rectangular sizes.

Add all six face areas.

Since \(1000\text{ cm}^3=1\) litre, \[ 277088.472\ldots\text{ cm}^3=277.088\ldots\text{ litres} \]

The cross-section is an L-shape.

Area of the lower rectangle:

Area of the upper rectangle:

So the cross-sectional area is

In the first \(1.5\) hours, the water level drops by \(20\) cm. This is the top rectangular part:

So the pumping rate is

Time to empty the whole container:

Starting at \(10{:}30\), ten hours later is

For the cuboid, \[ 12\times 5\times x=270 \]

The cylinder has radius \(4.5\text{ cm}\) and height \(9\text{ cm}\).

Also, \[ 1\text{ m}^3=100^3=1000000\text{ cm}^3 \]

Radius \(2x=9\): \[2\pi(9)^2=162\pi=508.9\ldots\]

The volume of a hemisphere is

For the original hemisphere \(H\),

Cancel \(\pi\):

The outer radius of the bowl is \(21\) cm.

The thickness is \(2\) cm, so the inner radius is

Total surface area of the bowl is:

- outer curved hemisphere - inner curved hemisphere - flat annular rim

Outer curved surface area:

Inner curved surface area:

Flat rim area:

Total:

Cuboid has a square cross-section of side *x* cm and length (15−4*x*) cm. Find the maximum value of the volume *V*.

Write *V* as a function of *x*: *V* = *x* ⋅ *x* ⋅ (15−4*x*) = *x*²(15−4*x*) = 15*x*² − 4*x*³

Differentiate: $$\\frac{dV}{dx} = 30x - 12x^2$$

Set $\\dfrac{dV}{dx} = 0$: $$30x - 12x^2 = 0 \\implies 6x(5-2x) = 0 \\implies x=0\\;\\text{ or }\\;x=\\tfrac{5}{2}=2.5$$ (*x* = 0 gives zero volume, so the meaningful solution is *x* = 2.5.)

Second derivative test to confirm maximum: $$\\frac{d^2V}{dx^2} = 30-24x\\;\\Bigg\|\_{x=2.5} = 30 - 60 = -30 \< 0 \\quad \\checkmark\\text{ (maximum)}$$

Calculate *V*_max: *V*_max = (2.5)² × (15−4×2.5) = 6.25 × 5 = 31.25 cm³

Calculus optimisation: always verify the nature of the stationary point using $\\dfrac{d^2V}{dx^2}$. Also check *x* \> 0 and 15 − 4*x* \> 0 (i.e. *x* \< 3.75) to ensure a valid cuboid.

Hemisphere H has radius *x* cm and volume 6174*π* cm³. A bowl is made by removing an inner hemisphere, leaving 2 cm thickness. Find total surface area in terms of *π*.

Find outer radius *x*. Volume of hemisphere $= \\dfrac{2}{3}\\pi r^3$: $$\\frac{2}{3}\\pi x^3 = 6174\\pi \\implies x^3 = \\frac{6174\\times3}{2} = 9261 \\implies x = \\sqrt\[3\]{9261} = 21\\text{ cm}$$

Find inner radius: *r*_inner = 21 − 2 = 19 cm

Three surfaces to add:

**1. Outer curved surface** (half-sphere, radius 21): *A*₁ = 2*π*(21)² = 2*π* × 441 = 882*π*

**2. Inner curved surface** (half-sphere, radius 19): *A*₂ = 2*π*(19)² = 2*π* × 361 = 722*π*

**3. Flat annular ring** at the top rim: *A*₃ = *π*(21)² − *π*(19)² = *π*(441−361) = 80*π*

Total surface area: *A*_total = 882*π* + 722*π* + 80*π* = 1684*π*

Three separate surfaces! Missing the flat annular ring is the most common error on this type of question.

The cone has radius \(5x\) and height \(6x\).

Volume of the cone:

The removed hemisphere has radius \(2x\).

Volume of the hemisphere:

The remaining solid has volume \(6948\pi\), so

Cancel \(\pi\):

The solid hemisphere removed has radius \(2x\).

Total surface area of a solid hemisphere is

So

The triangular cross-section is triangle \(DEF\).

It is right-angled at \(E\), with

and

So

and

Area of triangle \(DEF\):

The length of the prism is

Volume of the prism:

Let the radius of the smaller sphere be \(r\).

Let the radius of the larger sphere be \(R\).

Surface area of a sphere is

The surface area of sphere \(A\) is nine times the surface area of sphere \(B\), so

The difference between the volumes is \(117\pi\):

Cancel \(\pi\):

The cone has radius \(20\) cm.

The curved surface area of the cone is

So

Use Pythagoras to find the height of the cone:

Volume of the cone:

Volume of the hemisphere:

Total volume:

The volume is \(k\pi\), so

Find the height of the triangular cross-section: \[ h^2=11.2^2-7.4^2 \]

Area of the triangular cross-section: \[ \frac12(7.4)(8.407\ldots)=31.106\ldots \]

Volume of the prism: \[ 31.106\ldots\times 15=466.596\ldots \]

The curved surface area of a cone is

Here \(r=20\) and \(\pi r l=580\pi\), so

Use Pythagoras to find the height of the cone:

Volume of the cone:

Volume of the hemisphere:

Total volume:

Since the volume is \(k\pi\),

In triangle \(DBC\),

The circumference of the cylinder is \(DC\), so

Volume:

Jonty paints the four sides and the top, but not the base.

Top:

Two long sides:

Two short sides:

Total area:

Each tin covers \(15\text{ m}^2\), so the number of tins needed is

He must buy \(8\) tins.

After the \(10\%\) increase, each tin costs \(\pounds 26.95\). Before the increase:

Before the increase, \(8\) tins would cost

Since \(\pounds 196<\pounds 200\), Jonty is correct.

The sphere has diameter \(x\), so its radius is

The volume of the sphere is \(288\pi\text{ cm}^3\):

The rectangular base of the pyramid has dimensions \(x\) and \(2x\), so its area is

For the two faces with base \(x\), the slant height is

For the two faces with base \(2x\), the slant height is

Total surface area:

Substitute \(x=12\):

Correct to the nearest \(\text{cm}^2\):

Let the radius of the base of the cone be \(r\), and let the slant height be \(l\).

The volume of the cone is

The height is \(25\) cm, so

The slant height is found using Pythagoras:

When the sector is joined to form the cone, the arc length of the sector becomes the circumference of the cone base.

Cancel \(2\pi\):

The cross-section is a trapezium with parallel sides \(37\) cm and \(28\) cm, and perpendicular height \(20\) cm.

Since the trapezium is isosceles, the horizontal difference on each side is \[ \frac{37-28}{2}=4.5 \]

So \[ AB=CD=\sqrt{20^2+4.5^2}=20.5 \]

The perimeter of the cross-section is \[ 37+28+20.5+20.5=106 \]

The length of the prism is \(24\) cm, so the total surface area is \[ 2(650)+106(24)=3844 \]

For the angle between \(AF\) and the horizontal plane, the horizontal projection has length \[ \sqrt{24^2+4.5^2} \]

The vertical height is \(20\), so \[ \tan\theta=\frac{20}{\sqrt{24^2+4.5^2}} \]