Maths · 6 Mensuration

6.5 Compound shapes and parts of shapes

Before you start

The perimeter and area of a rectangle, triangle, parallelogram and trapezium, and splitting a compound shape into rectangles (6.2). Only the area of a triangle is on the formula list.
The circumference $2 π r$ and area $π r^{2}$ of a circle, and arc length and sector area as fractions of them (6.3).
The surface area and volume of a cuboid, prism, cylinder, sphere, pyramid and cone (6.4).
Pythagoras' theorem (7.1), to find a sloping edge or a slant height.
Lengths in similar shapes (5.3): corresponding lengths are in the same ratio.

Syllabus objectives covered (2)

Carry out calculations and solve problems involving perimeters and areas of: • compound shapes • parts of shapes.
Carry out calculations and solve problems involving surface areas and volumes of: • compound solids • parts of solids.

Part 1 of 6

See the pieces: add them, or take one away

A compound shape is built from shapes you already know: rectangles, triangles, trapezia, circles and parts of circles. You need no new formula. The skill is to see the pieces, and to find any length the question does not give you.
There are two ways in. Split the shape into pieces whose areas you can find, and add them. Or take away: start from one simple shape that covers the whole thing, and subtract the part that is missing. You used both with rectangles in 6.2. Here they reach triangles, circles and parts of circles.
Take a shape like the end of a house: a rectangle $10$ cm wide and $6$ cm tall, with an isosceles triangle of base $10$ cm and height $4$ cm on top. Split it along the top of the rectangle and add the two areas.
$10 \times 6 + \frac{1}{2} \times 10 \times 4 = 60 + 20 = 80$
graph diagram — not drawn yet
A pentagon shaped like the end of a house: a rectangle 10 cm wide and 6 cm tall with an isosceles triangle of base 10 cm and height 4 cm on top; the rectangle's top edge drawn as a dashed line inside the shape, the triangle's height drawn dashed from the apex down to the midpoint of that line, and each sloping edge labelled with a question mark.
Split along the dashed line for the area. For the perimeter the dashed line is inside the shape, and each sloping edge comes from Pythagoras.
So the area is $80$ cm². Notice that each piece uses its own lengths. The triangle's height is $4$ cm, measured from its own base. It is not $6 + 4 = 10$ cm, which is the height of the whole shape.
Take away is the natural method for a shaded region, where one shape is cut out of another. A square of side $10$ cm with a circle of radius $3$ cm cut from its middle has a shaded area of $100 - 9 π = 71.7$ cm² (3 s.f.).
$10^{2} - π \times 3^{2} = 100 - 9 π$
Choose the method with fewer pieces. The square with a hole is one subtraction. To split it you would need curved pieces that no formula covers.
Some lengths are not given. Find them from the ones that are: opposite sides of a rectangle are equal, a diameter is two radii, and a sloping edge may need Pythagoras. Write each found length on your sketch before you use it.
What the paper gives you: the area of a triangle, and the area and circumference of a circle, are on the formula list. The areas of a rectangle, a parallelogram and a trapezium are not, so you must recall them.

Part 2 of 6

Perimeter: walk round the outside edge only

The perimeter is the total length of the boundary. Start at one corner, go round the outside, and add each edge as you pass it, until you are back where you started.
Lines inside the shape are not part of the perimeter. That includes a line you drew to split the shape, and an edge where two pieces join. They help with the area. They are not on the boundary.
For the house shape, the boundary is the $10$ cm bottom, the two $6$ cm sides, and the two sloping edges of the triangle. The top of the rectangle is now inside the shape, so it does not count.
The sloping edges are not given. The triangle is isosceles, so its height meets the base at the midpoint, $5$ cm from each end. Each sloping edge is the hypotenuse of a right-angled triangle with sides $5$ cm and $4$ cm.
$\sqrt{5^{2} + 4^{2}} = \sqrt{41} = 6.403 \dots$
Add the five edges, and keep $\sqrt{41}$ unrounded in the calculator until the end. The perimeter is $34.8$ cm (3 s.f.).
$10 + 6 + 6 + 2 \sqrt{41} = 22 + 2 \sqrt{41} = 34.806 \dots$
Count the edges before you add. This shape has five. If your sum has four numbers, you have left one out.
Curved edges count too. A $20$ cm by $10$ cm rectangle with a semicircle on each $10$ cm end has a boundary of two straight $20$ cm edges and two semicircular arcs. The two $10$ cm ends of the rectangle are now inside the shape.
graph diagram — not drawn yet
A 20 cm by 10 cm rectangle with a semicircle of diameter 10 cm on each short end; the outside boundary drawn thick, and the two 10 cm ends of the rectangle drawn as thin dashed lines inside the shape.
The perimeter is the thick line: two straight 20 cm edges and two semicircular arcs. The dashed diameters are inside the shape.
Two semicircles with the same diameter make one full circle, so the two arcs together are $π \times 10 = 10 π$ . The perimeter is $40 + 10 π = 71.4$ cm (3 s.f.).
$P = 20 + 20 + 10 π = 40 + 10 π$
Its area is the rectangle plus one full circle of radius $5$ cm: $200 + 25 π = 279$ cm² (3 s.f.). For the area, the inside $10$ cm lines do no harm. For the perimeter, leave them out.

Part 3 of 6

Parts of a circle: take the right fraction

A semicircle is half a circle and a quarter circle is a quarter of one. A sector with angle $θ$ at the centre is $\frac{θ}{360}$ of the circle, as in 6.3. Each part has that fraction of the area and that fraction of the circumference.
For the area, multiply $π r^{2}$ by the fraction.
$semicircle: \frac{1}{2} π r^{2} quarter circle: \frac{1}{4} π r^{2}$
For the curved edge, the arc, multiply the circumference $2 π r$ by the same fraction. A semicircle's arc is $\frac{1}{2} \times 2 π r = π r$ .
The perimeter of the part on its own is the arc plus its straight edges. A semicircle has one straight edge, the diameter $2 r$ . A quarter circle has two, both radii, and so does any sector.
$P_{semicircle} = π r + 2 r P_{quarter} = \frac{1}{2} π r + 2 r$
Example: a quarter circle of radius $6$ cm. Its area is $\frac{1}{4} \times π \times 6^{2} = 9 π = 28.3$ cm². Its arc is $\frac{1}{4} \times 2 π \times 6 = 3 π$ , so its perimeter is $3 π + 6 + 6 = 21.4$ cm.
Before you use any formula, write $r = \dots$ . A semicircle drawn on a $10$ cm edge has diameter $10$ cm and radius $5$ cm. If you use $10$ as the radius, the area comes out four times too big.
In terms of π means leave $π$ as a symbol and give an exact answer, such as $9 π$ cm². Collect the $π$ terms together and the plain numbers together, but do not combine the two: $12 + 3 π$ is finished, and it is not $15 π$ .
When the question does not ask for $π$ , use the calculator's $π$ and round only the final answer, to 3 significant figures unless the question says otherwise.
Take away also gives a segment, the part of a circle between a chord and its arc: the sector minus the triangle. For the quarter circle of radius $6$ cm, the triangle has two $6$ cm sides at a right angle, so the segment's area is $9 π - 18 = 10.3$ cm².
$9 π - \frac{1}{2} \times 6 \times 6 = 9 π - 18$

Part 4 of 6

Compound solids: volumes add, hidden faces drop out

A compound solid is built from solids you know: a cone on a cylinder, a hemisphere on a cylinder, a prism-shaped roof on a cuboid. As with flat shapes, the work is in seeing the parts.
Volume is the easy half. Find the volume of each part and add. Joining two solids does not remove any space, so nothing is lost and nothing is counted twice. If a part is cut out, such as a hole drilled through a block, subtract its volume instead.
Surface area needs care. It is the area of every face on the outside. Where two parts join, the joining face is covered on both sides. It is inside the solid, so it does not count for either part.
graph diagram — not drawn yet
A cylinder of radius 4 cm and height 10 cm standing upright with a cone of radius 4 cm and slant height 5 cm on top; the circle where they meet drawn dashed and labelled 'hidden: not counted', and the base circle, the curved side of the cylinder and the curved surface of the cone each labelled.
Three surfaces are on the outside. The dashed circle where the cone meets the cylinder is covered on both sides, so it is not part of the surface area.
Take a cylinder of radius $4$ cm and height $10$ cm with a cone of radius $4$ cm and slant height $5$ cm on top. List each surface and decide whether it is outside. Base of the cylinder: outside, $π \times 4^{2} = 16 π$ . Curved side of the cylinder: outside, $2 π \times 4 \times 10 = 80 π$ .
Top of the cylinder and base of the cone: both hidden. Curved surface of the cone: outside, $π \times 4 \times 5 = 20 π$ . The total is $116 π = 364$ cm² (3 s.f.).
$16 π + 80 π + 20 π = 116 π$
Writing the list out like this shows the examiner each piece you counted, and it stops you adding a circle that is hidden.
The cone's volume needs its perpendicular height, not its slant height. By Pythagoras the height is $\sqrt{5^{2} - 4^{2}} = 3$ cm. The volume of the solid is $176 π = 553$ cm³ (3 s.f.).
$π \times 4^{2} \times 10 + \frac{1}{3} π \times 4^{2} \times 3 = 160 π + 16 π = 176 π$
A hole changes the surface in two ways. Drill a hole of radius $2$ cm straight through a cube of edge $10$ cm. Two circles of the cube's faces are gone, $- 2 \times 4 π$ . The inside of the hole is new surface, the curved surface of a cylinder, $+ 2 π \times 2 \times 10$ .
$600 - 8 π + 40 π = 600 + 32 π$
So the surface area rises to $600 + 32 π = 701$ cm² (3 s.f.), while the volume falls to $1000 - 40 π = 874$ cm³ (3 s.f.). A hole removes volume but can add surface.
What the paper gives you: the formula list has the curved surface areas of a cylinder and a cone, the surface area of a sphere, and the volumes of a prism, pyramid, cylinder, cone and sphere. It does not tell you which flat faces to include. That decision is yours, so show it.

Part 5 of 6

Parts of solids: hemispheres and half-cylinders

A hemisphere is half a sphere, so its volume is half of the sphere's $\frac{4}{3} π r^{3}$ .
$V = \frac{1}{2} \times \frac{4}{3} π r^{3} = \frac{2}{3} π r^{3}$
Its curved surface is half of the sphere's surface: $\frac{1}{2} \times 4 π r^{2} = 2 π r^{2}$ .
A solid hemisphere also has a flat circular face where the sphere was cut, of area $π r^{2}$ . So its total surface area is $2 π r^{2} + π r^{2} = 3 π r^{2}$ .
Read the wording. The outside of an open bowl is only the curved part, $2 π r^{2}$ . A hemisphere fixed flat onto another solid also shows only $2 π r^{2}$ , because its flat face is hidden. The total surface area of a solid hemisphere on its own is $3 π r^{2}$ .
The general rule: a cut makes a new flat face. When you work with part of a solid, add the face that the cut created.
A cylinder cut in half along its length is a half-cylinder. It is a prism, because its cross-section is the same semicircle all the way along. So its volume is the area of the semicircle times the length.
For radius $3$ cm and length $10$ cm, the volume is $45 π = 141$ cm³ (3 s.f.).
$\frac{1}{2} π \times 3^{2} \times 10 = 45 π$
Its surface has four faces. Half the curved surface: $\frac{1}{2} \times 2 π \times 3 \times 10 = 30 π$ . Two semicircular ends: $2 \times \frac{1}{2} π \times 3^{2} = 9 π$ . The flat rectangle the cut made, which is the diameter times the length: $6 \times 10 = 60$ .
$30 π + 9 π + 60 = 39 π + 60$
So the surface area is $39 π + 60 = 183$ cm² (3 s.f.). The rectangle is the face that is easy to leave out. Without it you get $39 π = 123$ cm², which is wrong.

Part 6 of 6

The frustum: a big cone minus a small cone

A frustum is what is left of a cone when a smaller cone is cut off its top, by a cut parallel to the base. A bucket and a lampshade are both frustums.
No frustum formula is on the formula list, and you do not need one. Treat the frustum as part of a solid: the full cone minus the small cone. Here $R$ and $H$ are the radius and height of the full cone, and $r$ and $h$ are those of the small cone.
$V_{frustum} = \frac{1}{3} π R^{2} H - \frac{1}{3} π r^{2} h$
graph diagram — not drawn yet
A side view of a cone of base radius R and height H, with a smaller cone of radius r and height h cut off at the top by a horizontal line parallel to the base; the small cone shaded and the frustum below it unshaded, with the right-angled triangle (h, r) inside the small cone and the right-angled triangle (H, R) inside the full cone both marked.
The frustum is the full cone minus the shaded small cone. The two right-angled triangles are similar, so r/R = h/H.
The small cone is the same shape as the full one, only smaller. The two cones are similar (5.3), so their radii and heights are in the same ratio.
$\frac{r}{R} = \frac{h}{H}$
Sometimes you are given only the frustum: its two radii and its own height. Then rebuild the full cone. Call the height of the missing small cone $x$ . The full cone's height is $x$ plus the frustum's height, and the ratio above gives an equation in $x$ . Worked example 4 does this.
The curved surface works the same way: the full cone's curved surface minus the small cone's, $π R L - π r l$ , where $L$ and $l$ are the two slant heights.
Slant heights are not always given. Find each one from its own cone's radius and height by Pythagoras, for example $L = \sqrt{R^{2} + H^{2}}$ .
Then add only the flat circles the question asks for. A solid frustum has two, $π R^{2}$ and $π r^{2}$ . An open bucket has only its base. A lampshade, open at both ends, has none.
Be careful with a cut halfway up. The small cone then has half the radius and half the height, so its volume is $\frac{1}{3} π {(\frac{R}{2})}^{2} (\frac{H}{2})$ , which is $\frac{1}{8}$ of the full cone. The frustum keeps $\frac{7}{8}$ of the volume, not half.

Worked examples

Example 1. A shape is made from a rectangle $12$ cm long and $8$ cm wide, with a semicircle on one of its $8$ cm sides. Calculate (i) the area and (ii) the perimeter of the shape. Give each answer correct to 3 significant figures.

Show the working

Find the radius first. The semicircle sits on an $8$ cm side, so that side is its diameter and $r = 4$ cm.
(i) The area is the rectangle plus half a circle.
$12 \times 8 + \frac{1}{2} π \times 4^{2} = 96 + 8 π$
Evaluate in one go and round at the end: $96 + 8 π = 121.13 \dots$ , so the area is $121$ cm².
(ii) Walk round the outside. The straight edges are $12$ , $8$ and $12$ . The fourth side of the rectangle, under the semicircle, is inside the shape, and the arc takes its place.
The arc is half the circumference of a circle of radius $4$ cm.
$\frac{1}{2} \times 2 π \times 4 = 4 π$
Add the outside edges, then round.
$12 + 8 + 12 + 4 π = 32 + 4 π = 44.566 \dots$
So the perimeter is $44.6$ cm. If you also added the $8$ cm line under the semicircle, you would get $52.6$ cm. That line is inside the shape, so that answer is wrong.

Answer: (i) $121$ cm² (exactly $96 + 8 π$ ) (ii) $44.6$ cm (exactly $32 + 4 π$ )

Example 2. $A B C D$ is a square of side $10$ cm. A quarter circle with centre $A$ and radius $10$ cm is drawn from $B$ to $D$ . The region inside the square but outside the quarter circle is shaded. Find the area and the perimeter of the shaded region in terms of $π$ , and then correct to 3 significant figures.

Show the working

The shaded region is the square with a piece taken away, so use take away: the square minus the quarter circle.
$10^{2} - \frac{1}{4} π \times 10^{2} = 100 - 25 π$
That is exact, so it is the area in terms of $π$ . As a decimal, $100 - 25 π = 21.46 \dots$ , which is $21.5$ cm².
For the perimeter, trace the boundary of the shaded region only. It runs from $B$ to $C$ ( $10$ cm), from $C$ to $D$ ( $10$ cm), and back from $D$ to $B$ along the arc. The sides $A B$ and $A D$ border the quarter circle, not the shaded region, so they are not on its boundary.
The arc is a quarter of the circumference of a circle of radius $10$ cm.
$\frac{1}{4} \times 2 π \times 10 = 5 π$
Add the three edges.
$P = 10 + 10 + 5 π = 20 + 5 π = 35.70 \dots$
So the perimeter is $20 + 5 π$ cm, which is $35.7$ cm. The $20$ and the $5 π$ are unlike terms, so $20 + 5 π$ is already the final exact form.

Answer: Area $100 - 25 π$ cm² $= 21.5$ cm²; perimeter $20 + 5 π$ cm $= 35.7$ cm.

Example 3. A solid is made from a cylinder of radius $3$ cm and height $10$ cm, with a hemisphere of radius $3$ cm fixed on top. Find (i) its volume and (ii) its total surface area. Give each answer in terms of $π$ , and then correct to 3 significant figures.

Show the working

(i) Add the volumes of the two parts. The hemisphere is half a sphere.
$π \times 3^{2} \times 10 + \frac{1}{2} \times \frac{4}{3} π \times 3^{3} = 90 π + 18 π = 108 π$
$108 π = 339.29 \dots$ , so the volume is $339$ cm³.
(ii) List each surface and decide whether it is outside. The base of the cylinder is outside: $π \times 3^{2} = 9 π$ . The curved side of the cylinder is outside: $2 π \times 3 \times 10 = 60 π$ .
The top of the cylinder and the flat face of the hemisphere press against each other, so both are hidden. Only the hemisphere's curved surface counts: $\frac{1}{2} \times 4 π \times 3^{2} = 18 π$ .
Add the three outside surfaces.
$9 π + 60 π + 18 π = 87 π$
$87 π = 273.318 \dots$ , so the surface area is $273$ cm².
If you had used $3 π r^{2}$ for the hemisphere, you would have added the hidden circle and got $96 π$ . A hemisphere on its own shows its flat face. A hemisphere fixed to a cylinder does not.

Answer: (i) $108 π$ cm³ $= 339$ cm³ (ii) $87 π$ cm² $= 273$ cm²

Example 4. A solid frustum has a top radius of $2$ cm, a base radius of $5$ cm and a height of $6$ cm. Calculate (i) its volume and (ii) its total surface area. Give your answers correct to 3 significant figures.

Show the working

Rebuild the full cone. Let the small cone that was removed have height $x$ cm. The full cone then has height $x + 6$ and radius $5$ . The small cone has height $x$ and radius $2$ .
The two cones are similar, so height divided by radius is the same for both.
$\frac{x}{2} = \frac{x + 6}{5}$
Multiply both sides by $10$ to clear the fractions: $5 x = 2 (x + 6)$ , so $5 x = 2 x + 12$ , $3 x = 12$ and $x = 4$ . The full cone has height $4 + 6 = 10$ cm.
(i) The volume is the full cone minus the small cone.
$\frac{1}{3} π \times 5^{2} \times 10 - \frac{1}{3} π \times 2^{2} \times 4 = \frac{250 π}{3} - \frac{16 π}{3} = 78 π$
$78 π = 245.04 \dots$ , so the volume is $245$ cm³.
(ii) Find the slant heights by Pythagoras. Full cone: $L = \sqrt{5^{2} + 10^{2}} = \sqrt{125}$ . Small cone: $l = \sqrt{2^{2} + 4^{2}} = \sqrt{20}$ . Keep them as square roots or full calculator values. Do not round them yet.
The curved surface of the frustum is the full cone's curved surface minus the small cone's.
$π \times 5 \times \sqrt{125} - π \times 2 \times \sqrt{20} = 147.521 \dots$
The frustum is solid, so add both circles: $π \times 5^{2} + π \times 2^{2} = 29 π = 91.106 \dots$
Add the curved surface and the two circles, then round.
$147.521 \dots + 91.106 \dots = 238.627 \dots$
So the total surface area is $239$ cm².

Answer: (i) $78 π$ cm³ $= 245$ cm³ (ii) $239$ cm²

Where people go wrong

Mistake: Adding the line where two shapes join into the perimeter, such as the side of a rectangle that a semicircle sits on.

Instead: Walk round the outside edge only. The joining line is inside the shape, and the arc takes its place.

Perimeter is the length of the boundary. In Worked example 1, the extra $8$ cm turns the correct $44.6$ cm into $52.6$ cm.

Mistake: Including the circle where two solids join in the surface area.

Instead: Leave out both the top of the lower solid and the base of the upper one.

Those two faces press against each other. Neither is on the outside of the solid, so neither is part of its surface.

Mistake: Giving the perimeter of a semicircle as $π r$ , or of a quarter circle as $\frac{1}{2} π r$ .

Instead: That is only the arc. Add the straight edges: $π r + 2 r$ for a semicircle and $\frac{1}{2} π r + 2 r$ for a quarter circle.

A part of a circle on its own is closed by its straight edges as well as its arc. When the part is joined to another shape, check which of those straight edges are now inside.

Mistake: Finding a frustum's volume as $\frac{1}{3} π (R^{2} - r^{2}) h$ , using the frustum's own height.

Instead: Find the full cone and the small cone, each with its own radius and height, and subtract their volumes.

The two cones have different heights, so you cannot take out a common height. In Worked example 4 the shortcut gives $\frac{1}{3} π (5^{2} - 2^{2}) \times 6 = 42 π$ , but the true volume is $78 π$ .

Mistake: Thinking that a cut halfway up a cone leaves half of its volume.

Instead: The small cone has half the radius and half the height, so it is $\frac{1}{8}$ of the full cone. The frustum keeps $\frac{7}{8}$ .

The volume uses the radius squared times the height: ${(\frac{1}{2})}^{2} \times \frac{1}{2} = \frac{1}{8}$ .

Check yourself

Find the area of a semicircle of radius $5$ cm. Give your answer in terms of $π$ .
Answer
$\frac{1}{2} \times π \times 5^{2} = \frac{25 π}{2} = 12.5 π$ cm².
A semicircle has a diameter of $10$ cm. Calculate its perimeter, correct to 3 significant figures.
Answer
The arc is $\frac{1}{2} \times π \times 10 = 5 π$ . Add the diameter: $5 π + 10 = 25.7$ cm.
A shape is made from a rectangle $8$ cm by $5$ cm with an isosceles triangle on one $8$ cm side. The triangle's height is $3$ cm. Find the area and the perimeter of the shape.
Answer
Area $8 \times 5 + \frac{1}{2} \times 8 \times 3 = 40 + 12 = 52$ cm². Each sloping edge is $\sqrt{4^{2} + 3^{2}} = 5$ cm, so the perimeter is $8 + 5 + 5 + 5 + 5 = 28$ cm. The $8$ cm side under the triangle is inside the shape.
A ring is the region between two circles with the same centre, of radii $7$ cm and $4$ cm. Find the area of the ring in terms of $π$ .
Answer
$π \times 7^{2} - π \times 4^{2} = 49 π - 16 π = 33 π$ cm².
A sector of a circle of radius $4$ cm has an angle of $90^{\circ}$ at the centre. Find the area of the segment between the arc and the chord that joins the ends of the arc, correct to 3 significant figures.
Answer
Sector $\frac{1}{4} \times π \times 4^{2} = 4 π$ . Triangle $\frac{1}{2} \times 4 \times 4 = 8$ . Segment $4 π - 8 = 4.57$ cm².
Find the volume of a hemisphere of radius $6$ cm, in terms of $π$ .
Answer
$\frac{2}{3} π \times 6^{3} = \frac{2}{3} π \times 216 = 144 π$ cm³.
Find the total surface area of a solid hemisphere of radius $6$ cm, in terms of $π$ .
Answer
Curved surface $2 π \times 6^{2} = 72 π$ , flat face $π \times 6^{2} = 36 π$ , total $108 π$ cm².
A cone stands on top of a cylinder with the same radius. Explain why the circle where they meet is not part of the surface area of the solid.
Answer
The top of the cylinder is covered by the cone, and the base of the cone is covered by the cylinder. Both faces are inside the solid, not on its outside, so neither is part of its surface.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A window is in the shape of a rectangle with a semicircle on top. The rectangle is $90$ cm wide and $150$ cm tall. The diameter of the semicircle is the whole $90$ cm top edge of the rectangle.

(a) Calculate the area of the window. Give your answer correct to 3 significant figures. 3 marks

Mark scheme

M1 $\frac{1}{2} \times π \times 45^{2}$ oe $(= 3180.8 \dots)$ seen
M1 $90 \times 150 +$ their semicircle area
A1 awrt $16 700$ (cm²)

(b) Calculate the perimeter of the window. 3 marks

Mark scheme

M1 $\frac{1}{2} \times π \times 90$ oe, e.g. $45 π$ $(= 141.3 \dots)$ seen
M1 $150 + 90 + 150 +$ their arc oe, with no $90$ cm edge under the semicircle
A1 awrt $531$ (cm)

Question 2. A solid toy is made from a cone and a hemisphere. The base of the cone fits exactly against the flat face of the hemisphere. The cone has base radius $5$ cm, perpendicular height $12$ cm and slant height $13$ cm. The hemisphere has radius $5$ cm.

(a) Find the volume of the toy. Give your answer in terms of $π$ . 3 marks No calculator

Mark scheme

M1 $\frac{1}{3} \times π \times 5^{2} \times 12$ $(= 100 π)$
M1 $\frac{1}{2} \times \frac{4}{3} \times π \times 5^{3}$ oe, e.g. $\frac{2}{3} π \times 5^{3}$ $(= \frac{250 π}{3})$
A1 $\frac{550 π}{3}$ oe, e.g. $183 \frac{1}{3} π$

(b) Find the total surface area of the toy. Give your answer in terms of $π$ . 3 marks No calculator

Mark scheme

M1 $π \times 5 \times 13$ $(= 65 π)$
M1 $\frac{1}{2} \times 4 \times π \times 5^{2}$ oe, e.g. $2 π \times 5^{2}$ $(= 50 π)$
A1 $115 π$ cao, with no area of the joining circle included

Question 3. A frustum is made by removing a small cone of height $4$ cm from the top of a solid cone of base radius $9$ cm and height $12$ cm. The cut is parallel to the base.

(a) Show that the radius of the small cone is $3$ cm. 2 marks No calculator

Mark scheme

M1 $\frac{r}{9} = \frac{4}{12}$ oe, or scale factor $\frac{1}{3}$ seen
A1 $r = 3$ from correct working; nfww

(b) Calculate the volume of the frustum. 3 marks

Mark scheme

M1 $\frac{1}{3} \times π \times 9^{2} \times 12$ $(= 324 π)$ or $\frac{1}{3} \times π \times 3^{2} \times 4$ $(= 12 π)$
M1 their $324 π -$ their $12 π$ , or $\frac{26}{27} \times$ their $324 π$ oe
A1 awrt $980$ (cm³), or $312 π$

(c) Calculate the curved surface area of the frustum. 3 marks

Mark scheme

M1 slant height $\sqrt{9^{2} + 12^{2}} = 15$ or $\sqrt{3^{2} + 4^{2}} = 5$ , or frustum slant edge $\sqrt{6^{2} + 8^{2}} = 10$
M1 $π \times 9 \times 15 - π \times 3 \times 5$ oe, or $π \times (9 + 3) \times 10$ ; FT their slant heights
A1 awrt $377$ (cm²), or $120 π$

How this is examined

The volume or surface area of a frustum, as a full cone minus a small cone: a Calculate on a frustum, which may be split into parts: a Show that for a missing radius or height from the similar cones, then the volume, then the surface area For the volume: a method mark for the cone volumes, each written with its own radius and height, e.g. $\frac{1}{3} π \times 9^{2} \times 12$ and $\frac{1}{3} π \times 3^{2} \times 4$ ; a method mark for big minus small; then the accuracy mark. For the curved surface: a slant height by Pythagoras, then $π R L - π r l$ . For a missing radius, write the ratio, e.g. $\frac{r}{9} = \frac{4}{12}$ , and do not just state $r = 3$ .
Deciding which faces of a compound solid are on the outside: a Find or Calculate for the total surface area of two solids joined together, which may ask for the answer in terms of π A method mark for each outside surface, written separately, e.g. cone $π \times 5 \times 13$ and hemisphere $2 π \times 5^{2}$ ; the accuracy mark for their total. If you add the hidden joining circle, the total is wrong and the accuracy mark is not earned.
Adding the volumes of the parts of a compound solid: a Find or Calculate after the measurements of each part are given, e.g. a cone on a hemisphere or a hemisphere on a cylinder A method mark for each part's volume with the right formula: $\frac{1}{3} π r^{2} h$ with the perpendicular height, not the slant height, and half of $\frac{4}{3} π r^{3}$ for a hemisphere. The accuracy mark for the sum.
The area of a compound shape or shaded region, by splitting or taking away: a Calculate on a shape such as a rectangle with a semicircle on one side, or a square with a circle or a quarter circle removed A method mark for the area of the curved part with the radius, not the diameter, e.g. $\frac{1}{2} \times π \times 45^{2}$ for a semicircle on a $90$ cm edge; a method mark for the right combination, added or subtracted; the accuracy mark for the answer to 3 significant figures.
The perimeter of a shape with a curved edge: the arc plus the outside straight edges: a Calculate straight after the area of the same shape A method mark for the arc as a fraction of the circumference, e.g. $\frac{1}{2} \times π \times 90$ ; a method mark for the arc added to the outside straight edges only, with no line that is inside the shape; the accuracy mark for the total.
Giving an exact answer in terms of π: the instruction 'Give your answer in terms of π' at the end of a Find, on a non-calculator paper or a calculator one A final answer with $π$ left as a symbol and like terms collected, e.g. $115 π$ , $\frac{550 π}{3}$ or $20 + 5 π$ . A decimal such as $361$ is not in terms of $π$ , so it does not answer the question as asked.

Words to know

Compound shape: A flat shape made by joining simple shapes, such as rectangles, triangles and parts of circles.
Compound solid: A solid made by joining simple solids, such as a cone on a cylinder.
Arc: Part of the circumference of a circle.
Sector: The part of a circle between two radii and the arc that joins them. A sector with angle $θ$ is $\frac{θ}{360}$ of the circle.
Segment: The part of a circle between a chord and the arc it cuts off: the sector minus the triangle.
Hemisphere: Half a sphere. Volume $\frac{2}{3} π r^{3}$ and curved surface $2 π r^{2}$ . A solid hemisphere also has a flat circular face of area $π r^{2}$ .
Frustum: The part of a cone that is left when a smaller cone is cut off its top, parallel to the base.
Slant height: The distance from the vertex of a cone to the edge of its base, along the curved surface: $l = \sqrt{r^{2} + h^{2}}$ .
In terms of π: An exact answer with $π$ left as a symbol, such as $115 π$ or $20 + 5 π$ .

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