Hard GCSE Volume and surface area of prisms Questions

Realise that the amount of paper is the surface area.

The paper covers the outside of the box, so the area of paper needed is the total surface area of the cuboid.

Work out the three different face areas.

The three kinds of face have areas $240$, $200$ and $120$ square centimetres.

Add the three face areas and double.

Each face appears twice, so the box needs $1120$ $\text{cm}^2$ of paper.

Multiply the area by the cost of one square centimetre.

Each square centimetre costs $2$ pence.

Carry out the multiplication.

The paper costs $2240$ pence.

State the units of the answer.

Two lengths in centimetres are multiplied together, so an area is measured in square centimetres, $\text{cm}^2$.

Check the area against the net.

Adding the six rectangles of the net one at a time gives the same $1120$ $\text{cm}^2$.

Check the cost by dividing back.

Dividing the cost by $2$ pence per square centimetre returns the area $1120$ $\text{cm}^2$.

Convert the cost into pounds for a sanity check.

That is about 22.4 pounds, a believable price for a sheet of paper this size.

Note the common slip of using the volume.

The volume of the box is $2400$ $\text{cm}^3$, but paper covers a SURFACE, so it is the area, not the volume, that is needed.

Note that no paper is wasted in this model.

The calculation assumes the paper is cut to fit exactly, with no overlaps and no waste, which is what an exam question of this kind intends.

Identify the largest face.

The largest face is $240$ $\text{cm}^2$, so no single sheet smaller than that could cover it.

Check the surface area is bigger than the largest face.

The whole surface must be far bigger than any one face, and it is.

State the cost per face as a check.

Every square centimetre of the $1120$ needed costs $2$ pence, giving $2240$ pence in all.

State the total cost of the paper.

The paper costs $2240$ pence.

Realise that the amount of paper is the surface area.

The paper covers the outside of the box, so the area of paper needed is the total surface area of the cuboid.

Work out the three different face areas.

The three kinds of face have areas $150$, $120$ and $80$ square centimetres.

Add the three face areas and double.

Each face appears twice, so the box needs $700$ $\text{cm}^2$ of paper.

Multiply the area by the cost of one square centimetre.

Each square centimetre costs $3$ pence.

Carry out the multiplication.

The paper costs $2100$ pence.

State the units of the answer.

Two lengths in centimetres are multiplied together, so an area is measured in square centimetres, $\text{cm}^2$.

Check the area against the net.

Adding the six rectangles of the net one at a time gives the same $700$ $\text{cm}^2$.

Check the cost by dividing back.

Dividing the cost by $3$ pence per square centimetre returns the area $700$ $\text{cm}^2$.

Convert the cost into pounds for a sanity check.

That is about 21 pounds, a believable price for a sheet of paper this size.

Note the common slip of using the volume.

The volume of the box is $1200$ $\text{cm}^3$, but paper covers a SURFACE, so it is the area, not the volume, that is needed.

Note that no paper is wasted in this model.

The calculation assumes the paper is cut to fit exactly, with no overlaps and no waste, which is what an exam question of this kind intends.

Identify the largest face.

The largest face is $150$ $\text{cm}^2$, so no single sheet smaller than that could cover it.

Check the surface area is bigger than the largest face.

The whole surface must be far bigger than any one face, and it is.

State the cost per face as a check.

Every square centimetre of the $700$ needed costs $3$ pence, giving $2100$ pence in all.

State the total cost of the paper.

The paper costs $2100$ pence.

Identify the three parts of the surface.

The net of a cylinder is a rectangle for the curved surface together with two circles, one for each end.

Work out the area of the two circular ends.

Each end is a circle of area $\pi r^2 = 49\pi$ $\text{cm}^2$, so the two together cover $98\pi$ $\text{cm}^2$.

Work out the curved surface area.

The curved surface unrolls into a rectangle $14\pi$ cm wide and $11$ cm tall, so its area is $154\pi$ $\text{cm}^2$.

Add the curved surface to the two ends.

The total surface area is $252\pi$ $\text{cm}^2$.

Recall the area of a circle.

The flat cross-section of a cylinder is a circle, and the area of a circle of radius $r$ is $\pi r^2$.

Work out the area of one circular end on its own.

Each end has area $49\pi$ $\text{cm}^2$.

Recall the circumference of a circle.

The curved surface of a cylinder unrolls into a rectangle whose width is the circumference of the circular end, $2 \pi r$.

Check the width of the unrolled rectangle.

The rectangle has to wrap right round the circular end, so its width is the circumference $14\pi$ cm.

State the units of the answer.

Two lengths in centimetres are multiplied together, so an area is measured in square centimetres, $\text{cm}^2$.

Check against the standard formula.

The formula $2 \pi r^2 + 2 \pi r h$ gives the same $252\pi$.

Check by factorising the formula.

Factorising gives $2 \pi r (r + h)$, and $14 \times 18 = 252$, which agrees.

Leave the answer in terms of pi.

Leaving $\pi$ in the answer keeps it exact; replacing it by a decimal would only ever be an approximation.

Estimate the answer as a decimal.

Taking $\pi$ as about $3.14$ gives roughly $791.28$ square centimetres.

Compare the curved surface with the ends.

The curved surface contributes $154\pi$ and the ends only $98\pi$, so most of the surface is the side.

State the total surface area of the cylinder.

The total surface area is $252\pi$ $\text{cm}^2$.

Identify the five faces of the prism.

A triangular prism has two triangular ends and three rectangular faces, one for each side of the triangle.

Work out the area of one triangular end.

Half the base times the height gives $54$ $\text{cm}^2$.

Work out the perimeter of the triangle.

The three rectangular faces together form one long rectangle whose width is the perimeter of the triangle, $36$ cm.

Work out the area of the three rectangular faces together.

That long rectangle is $36$ cm by $20$ cm, so its area is $720$ $\text{cm}^2$.

Work out the area of the two triangular ends.

The two ends are identical, so together they cover $108$ $\text{cm}^2$.

Add the ends to the rectangular faces.

The total surface area is $828$ $\text{cm}^2$.

State the units of the answer.

Two lengths in centimetres are multiplied together, so an area is measured in square centimetres, $\text{cm}^2$.

Check the hypotenuse with Pythagoras.

The three sides really do make a right-angled triangle, so the $15$ cm side is the hypotenuse and the $9$ cm and $12$ cm sides are the ones at right angles.

List the three rectangles separately.

Each side of the triangle gives one rectangle of length $20$ cm.

Add the three rectangles to check the long rectangle.

They total $720$ $\text{cm}^2$, matching the perimeter method.

Add all five faces one by one as a final check.

Counting every face separately gives the same total, $828$ $\text{cm}^2$.

Contrast the surface area with the volume.

The volume uses the cross-sectional AREA times the length; the surface area uses the PERIMETER times the length plus the two ends. Mixing the two up is the commonest mistake here.

Check that the rectangular faces dominate.

The three rectangles cover $720$ $\text{cm}^2$ against $108$ $\text{cm}^2$ for the two ends, which fits a prism that is long compared with its cross-section.

Check the answer is bigger than the largest face.

No single face can be as big as the whole surface, and the largest one here is $300$ $\text{cm}^2$.

State the total surface area of the prism.

The total surface area is $828$ $\text{cm}^2$.

Recall the volume of a cylinder.

The volume of a cylinder is the area of its circular cross-section multiplied by its height.

Put the known values into the rule.

The volume is $768\pi$ $\text{cm}^3$ and the height is $12$ cm.

Divide both sides by pi.

Every term carries a factor of $\pi$, so it cancels straight away and the work stays exact.

Divide both sides by the height.

This leaves $r^2 = 64$.

Take the square root.

The radius is the positive square root, $8$ cm.

Reject the negative square root.

A radius is a length, so only the positive root $8$ makes sense here.

Check by working the volume forwards.

Substituting the radius back gives $768\pi$ $\text{cm}^3$, the volume we were given.

Work out the area of the circular end.

The cross-section has area $64\pi$ $\text{cm}^2$.

Check the cross-sectional area against the volume.

Cross-sectional area times height returns the volume, as it must for any prism.

Note the common slip of forgetting to square root.

Stopping at $64$ gives the SQUARE of the radius, not the radius. The square root step is essential.

Work out the diameter as well.

The cylinder is $16$ cm across.

Work out the circumference of the end.

The circular end has circumference $16\pi$ cm.

Estimate the volume as a decimal.

The cylinder holds about $2411.52$ cubic centimetres, which is a sensible size for a radius of $8$ cm and a height of $12$ cm.

Check the units of the answer.

The square root of an area is a length, so the radius is in centimetres.

State the radius of the cylinder.

The radius of the cylinder is $8$ cm.