Worked example · Easy
A rectangular field
Find the perimeter of a \(12\text{ m}\) by \(7\text{ m}\) field.
Add all four edges: \(P=12+7+12+7=\boxed{38\text{ m}}\).
Chapter 40 · Completely rewritten
Mensuration
Clear explanations, original worked examples and a full graded exercise set, mapped to the textbook.
What you will learn
Measure and convert lengths, areas and volumes; calculate perimeters, areas, arc lengths, surface areas and volumes for simple and compound figures; select and rearrange relevant formulas.
Begin with the easy examples, then attempt the medium and exam-style examples before using the hidden solutions.
1
Subtopic 1
The perimeter of a 2-D shape
Perimeter means the total distance round the outside of a 2-D shape. Measure only the outer boundary. For an L-shape, use the known horizontal and vertical totals to recover any missing side lengths before adding.
Every length must be in the same unit, and a perimeter answer has length units, such as \(\text{cm}\), not \(\text{cm}^2\).
Essential method
\(P=\text{sum of all outside side lengths}\). For a rectangle, \(P=2(l+w)\). A missing stepped-edge length often comes from subtraction.
Exam tip: Do not count shared internal edges or add centimetres to metres before converting.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Medium
A missing side
An L-shape has total width \(14\text{ cm}\), and its shorter horizontal part is \(9\text{ cm}\). Find the missing horizontal step.
The horizontal parts must total the full width: \(14-9=\boxed{5\text{ cm}}\).
Worked example · Exam-style
Mixed units
A rectangular noticeboard is \(1.5\text{ m}\) wide and \(80\text{ cm}\) high. Find the total edging in centimetres.
Convert \(1.5\text{ m}=150\text{ cm}\); perimeter \(=2(150+80)=\boxed{460\text{ cm}}\).
2
Subtopic 2
The area of a rectangle
Area measures the space inside a shape. Multiply two perpendicular dimensions; if the measurements are in centimetres, the answer is in square centimetres. Work backwards by dividing a known area by one side.
Essential method
\(A=lw\), so \(l=A/w\) and \(w=A/l\).
Exam tip: Always square the unit, not the number.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Floor tiles
Find the area of a floor measuring \(6\text{ m}\) by \(4\text{ m}\).
\(A=6\times4=\boxed{24\text{ m}^2}\).
Worked example · Medium
Find a missing dimension
A poster has area \(180\text{ cm}^2\) and width \(12\text{ cm}\). Find its length.
\(l=A/w=180/12=\boxed{15\text{ cm}}\).
Worked example · Exam-style
Mixed units and cost
A rectangular surface is \(1.2\text{ m}\) by \(75\text{ cm}\). Material costs $20 per \(\text{m}^2\). Find the material cost.
\(75\text{ cm}=0.75\text{ m}\), so \(A=1.2\times0.75=0.9\text{ m}^2\); cost \(=0.9\times20=\boxed{\$18}\).
3
Subtopic 3
The area of a triangle
Any triangle is half of a parallelogram with the same base and perpendicular height. The height may fall outside an obtuse triangle. A sloping side is not automatically the height.
Essential method
\(A=\tfrac12 bh\), where \(h\) is measured at \(90^\circ\) to the chosen base.
Exam tip: Use a perpendicular height, never simply another edge.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Right triangle
A right triangle has base \(8\text{ cm}\) and perpendicular height \(5\text{ cm}\). Find the area.
\(A=\tfrac12\times8\times5=\boxed{20\text{ cm}^2}\).
Worked example · Medium
Find the height
A triangle has area \(42\text{ cm}^2\) and base \(12\text{ cm}\). Find the perpendicular height.
\(42=\tfrac12(12)h=6h\), so \(h=\boxed{7\text{ cm}}\).
Worked example · Exam-style
Large triangle
A triangular park has a base of \(0.3\text{ km}\) and perpendicular height \(240\text{ m}\). Calculate its area in \(\text{m}^2\).
Convert \(0.3\text{ km}=300\text{ m}\). \(A=\tfrac12(300)(240)=\boxed{36\,000\text{ m}^2}\).
4
Subtopic 4
The area of a parallelogram
Move the triangular corner of a parallelogram to the other side and it forms a rectangle. This explains why we multiply the base by the perpendicular height rather than the sloping edge.
Essential method
\(A=bh\).
Exam tip: Check the given height really meets the base at a right angle.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Simple area
A parallelogram has base \(9\text{ cm}\) and perpendicular height \(4\text{ cm}\).
\(A=9\times4=\boxed{36\text{ cm}^2}\).
Worked example · Medium
Solve for height
Its area is \(84\text{ cm}^2\), with base \(12\text{ cm}\).
\(h=84/12=\boxed{7\text{ cm}}\).
Worked example · Exam-style
Sloping-side trap
A parallelogram has base \(14\text{ m}\), sloping edge \(8\text{ m}\), and perpendicular height \(6\text{ m}\). Find its area.
The sloping \(8\text{ m}\) is irrelevant. \(A=14\times6=\boxed{84\text{ m}^2}\).
5
Subtopic 5
The area of a trapezium
A trapezium has one pair of parallel sides. Its area is the average length of these parallel sides, multiplied by the perpendicular distance between them.
Essential method
\(A=\tfrac12(a+b)h\), where \(a,b\) are the parallel sides.
Exam tip: Do not use the sloping side in place of the perpendicular height.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Simple trapezium
The parallel sides are 8 cm and 12 cm, and the height is 5 cm.
\(A=\tfrac12(8+12)\times5=\boxed{50\text{ cm}^2}\).
Worked example · Medium
Missing height
A trapezium has area 72 cm² and parallel sides 10 cm and 14 cm.
\(72=\tfrac12(10+14)h=12h\), so \(h=\boxed{6\text{ cm}}\).
Worked example · Exam-style
Road cross-section
A trapezoidal section has widths 4.4 m and 7.6 m and depth 1.5 m. Find the cross-sectional area.
\(A=\tfrac12(4.4+7.6)\times1.5=\boxed{9\text{ m}^2}\).
6
Subtopic 6
The area of shapes made from rectangles and triangles
Divide a compound figure into rectangles, triangles or other familiar shapes; find the parts and add. Alternatively, enclose the shape in a large rectangle and subtract the area of each gap. Both approaches should agree.
Essential method
\(A_{\mathrm{compound}}=\sum A_{\mathrm{pieces}}\), or \(A_{\mathrm{outer}}-\sum A_{\mathrm{removed}}\).
Exam tip: Use a sketch to ensure pieces do not overlap or leave gaps.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
L-shaped floor
A 10 by 8 cm rectangle has a 4 by 3 cm corner removed. Find its area.
Outer area \(=10\times8=80\). Gap \(=4\times3=12\). Remaining \(=\boxed{68\text{ cm}^2}\).
Worked example · Medium
Triangle on a rectangle
A rectangle of area 36 cm² has a triangle of base 9 cm and height 4 cm attached without overlap.
Triangle area \(=\tfrac12(9)(4)=18\). Total \(=36+18=\boxed{54\text{ cm}^2}\).
Worked example · Exam-style
Cut-out patio
A patio is 12 m by 9 m, but a triangular region with base 6 m and height 5 m is unpaved. Find the paved area.
\(12(9)-\tfrac12(6)(5)=108-15=\boxed{93\text{ m}^2}\).
7
Subtopic 7
The circumference of a circle
A circle’s circumference is the distance all the way round. Its diameter is twice the radius. Circumference is a length, and \(\pi\) should be kept on the calculator until the end.
Essential method
\(C=2\pi r=\pi d\). To find the diameter from circumference, \(d=C/\pi\).
Exam tip: Check whether a figure gives diameter or radius.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Radius known
A wheel has radius \(7\text{ cm}\). Find its circumference in terms of \(\pi\).
\(C=2\pi(7)=\boxed{14\pi\text{ cm}}\).
Worked example · Medium
Diameter known
A circular track has diameter \(24\text{ m}\). Find its circumference to 1 decimal place.
\(C=\pi(24)=\boxed{75.4\text{ m}}\) (1 d.p.).
Worked example · Exam-style
Round a track
A wheel moves 1.5 m per full revolution. Estimate the number of revolutions in a 120 m journey.
Each revolution covers 1.5 m, so \(120/1.5=\boxed{80}\) revolutions.
8
Subtopic 8
The area of a circle
A circle’s area is proportional to the square of its radius. A ring (annulus) has the area of a large disk minus the small disk. Never use the circumference formula to calculate area.
Essential method
\(A=\pi r^2\), and \(A_{\mathrm{ring}}=\pi(R^2-r^2)\).
Exam tip: Convert diameter to radius before squaring.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Find the disk area
Find the area of a circle of radius 5 cm, in terms of \(\pi\).
\(A=\pi(5)^2=\boxed{25\pi\text{ cm}^2}\).
Worked example · Medium
Use the diameter
A circle has diameter 18 cm. Find its area.
Radius \(=9\text{ cm}\). Area \(=81\pi\approx\boxed{254.5\text{ cm}^2}\).
Worked example · Exam-style
Annular path
A pond of radius 3 m has a surrounding path 1 m wide. Find the path area.
Outer radius \(4\text{ m}\); path area \(=\pi(4^2-3^2)=\boxed{7\pi\text{ m}^2}\).
9
Subtopic 9
Arc length and sector area
A sector is a fraction of a whole circle. Its angle \(\theta\) must be measured at the centre. The fraction of a complete turn is \(\theta/360\). Its perimeter includes both radii as well as the curved arc.
Essential method
\(L=\frac{\theta}{360}\,2\pi r\), \(A=\frac{\theta}{360}\,\pi r^2\), \(P_{\rm sector}=2r+L\).
Exam tip: For a major sector use the major angle (or subtract the minor sector from a complete circle).
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Quarter-circle arc
Find the arc length for \(r=6\text{ cm},\theta=90^\circ\).
\(L=(90/360)2\pi(6)=\boxed{3\pi\text{ cm}}\).
Worked example · Medium
Sector area
Find the area of a \(60^\circ\) sector of radius 9 cm.
\(A=(60/360)\pi(9^2)=\boxed{13.5\pi\text{ cm}^2}\).
Worked example · Exam-style
Sector perimeter
A \(135^\circ\) sector has radius 8 cm. Find its perimeter.
Arc \(=(135/360)2\pi(8)=6\pi\). Add two radii: \(P=16+6\pi\approx\boxed{34.8\text{ cm}}\).
10
Subtopic 10
The volume of a prism
A prism has the same cross-section all along its length. This includes cuboids and cylinders. Find the area of the cross-section, then multiply by the perpendicular length of the prism.
Essential method
\(V=A_{\rm cross-section}\times l\). For a cylinder, \(V=\pi r^2h\).
Exam tip: Volume has cubic units. For cylinders, square the radius, not the height.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Cuboid
Find the volume of a 6 by 4 by 3 cm cuboid.
\(V=6\times4\times3=\boxed{72\text{ cm}^3}\).
Worked example · Medium
Triangular prism
A prism has triangular cross-section base 8 cm and height 5 cm, and length 12 cm.
Cross-section \(=\tfrac12(8)(5)=20\text{ cm}^2\). Volume \(=20(12)=\boxed{240\text{ cm}^3}\).
Worked example · Exam-style
Cylinder capacity
A cylindrical tank has radius 0.4 m and height 1.2 m. Find its capacity in litres, to the nearest litre.
\(V=\pi(0.4)^2(1.2)=0.192\pi\text{ m}^3\). Since \(1\text{ m}^3=1000\text{ L}\), capacity \(=192\pi\approx\boxed{603\text{ L}}\).
11
Subtopic 11
The surface area of a prism
Surface area is the sum of the exposed faces. For a closed prism, include both congruent end faces as well as every side face. For a cylinder, the curved surface unwraps to a rectangle of width equal to the circumference.
Essential method
For a closed prism, \(S=2A_{\rm end}+P_{\rm cross-section}\,l\). Closed cylinder: \(S=2\pi rh+2\pi r^2\).
Exam tip: Exclude faces that are not exposed, such as the missing top of an open container.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Cuboid total area
A cuboid measures 2 cm by 3 cm by 5 cm. Find its total surface area.
\(S=2(2\times3+3\times5+2\times5)=2(6+15+10)=\boxed{62\text{ cm}^2}\).
Worked example · Medium
Cylinder curved area
A cylinder has radius 4 cm and height 10 cm. Find curved surface area.
\(S_{\rm curved}=2\pi(4)(10)=\boxed{80\pi\text{ cm}^2}\).
Worked example · Exam-style
Open cylindrical tin
A cylindrical tin without a lid has radius 5 cm and height 12 cm. Find total metal area.
Curved area \(=2\pi(5)(12)=120\pi\); base \(=25\pi\); total \(=\boxed{145\pi\text{ cm}^2}\).
12
Subtopic 12
The volume of a pyramid, a cone and a sphere
A pyramid or cone occupies one third of the volume of a prism or cylinder having the same base area and perpendicular height. Sphere volume depends on the cube of its radius.
Essential method
\(V_{\rm pyramid}=\tfrac13 A_{\rm base}h\); \(V_{\rm cone}=\tfrac13\pi r^2h\); \(V_{\rm sphere}=\tfrac43\pi r^3\).
Exam tip: For cones and pyramids use the perpendicular vertical height, not the slant length.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Square pyramid
A pyramid has base 6 cm by 6 cm and perpendicular height 9 cm.
\(V=\tfrac13(6\times6)(9)=\boxed{108\text{ cm}^3}\).
Worked example · Medium
Cone
A cone has radius 3 cm and height 8 cm. Find its volume in terms of \(\pi\).
\(V=\tfrac13\pi(3^2)(8)=\boxed{24\pi\text{ cm}^3}\).
Worked example · Exam-style
Sphere from diameter
A solid ball has diameter 12 cm. Find its volume in terms of \(\pi\).
Radius \(6\text{ cm}\); \(V=\tfrac43\pi(6^3)=\boxed{288\pi\text{ cm}^3}\).
13
Subtopic 13
The surface area of a pyramid, a cone and a sphere
For a pyramid, add the base and all triangular faces. For a cone, the curved area uses the slant height \(l\); if only perpendicular height \(h\) is given, find \(l\) with Pythagoras. A sphere has one continuous curved surface.
Essential method
\(S_{\rm cone,curved}=\pi rl\), \(S_{\rm cone,total}=\pi rl+\pi r^2\), \(l=\sqrt{r^2+h^2}\); \(S_{\rm sphere}=4\pi r^2\).
Exam tip: For total cone area include the base unless the question says “curved” or “open”.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Sphere
Find the surface area of a sphere with radius 5 cm.
\(S=4\pi(5^2)=\boxed{100\pi\text{ cm}^2}\).
Worked example · Medium
Curved cone surface
A cone has radius 4 cm and slant height 9 cm. Find its curved area.
\(S=\pi(4)(9)=\boxed{36\pi\text{ cm}^2}\).
Worked example · Exam-style
Cone height given
A closed cone has base radius 5 cm and height 12 cm. Find total surface area.
\(l=\sqrt{5^2+12^2}=13\). Curved area \(=65\pi\), base \(=25\pi\), so \(\boxed{90\pi\text{ cm}^2}\).
14
Subtopic 14
The area and volume of compound shapes
For solids built by joining or removing simple solids, add or subtract volumes. For surface area, include only the outer faces left visible. A frustum can be calculated by subtracting the smaller similar cone from the original; similarity finds missing radii or heights.
A circular segment is the region between a chord and its arc: its area is the sector area minus the central triangle for a minor segment.
Essential method
\(V_{\rm frustum}=\tfrac13\pi R^2H-\tfrac13\pi r^2h\); \(A_{\rm segment}=\frac{\theta}{360}\pi r^2-\frac12r^2\sin\theta\). Segment perimeter \(=\text{arc length}+\text{chord length}\).
Exam tip: Use consistent length units, account for all exposed surfaces, and round only the final value.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Solid made from two blocks
A solid contains two non-overlapping cuboids of volumes 120 and 75 cm³. Find its volume.
Add volumes: \(120+75=\boxed{195\text{ cm}^3}\).
Worked example · Medium
Cylinder with drilled hole
A cylinder of radius 5 cm and height 10 cm has a cylindrical hole radius 2 cm drilled all the way through. Find remaining volume.
\(V=\pi(5^2)(10)-\pi(2^2)(10)=(250-40)\pi=\boxed{210\pi\text{ cm}^3}\).
Worked example · Exam-style
Conical frustum
A cone has height 12 cm and radius 6 cm. A smaller similar cone of height 4 cm is removed from the tip. Find the remaining volume.
By similarity, small radius \(=6(4/12)=2\text{ cm}\). \(V=\tfrac13\pi(6^2)(12)-\tfrac13\pi(2^2)(4)=144\pi-\tfrac{16\pi}{3}=\boxed{\tfrac{416\pi}{3}\text{ cm}^3}\).
Chapter 40 — Student exercises
Questions are arranged by textbook subtopic. Within each section they start with confidence-building tasks and finish with extended examination-style reasoning. Try each question before opening its solution.
The perimeter of a 2-D shape
1Question 1Foundation
A square has side \(8\text{ cm}\). Find its perimeter.
Show Answer
Step-by-step solution: Four equal sides: \(4\times8=\boxed{32\text{ cm}}\).
2Question 2Foundation
A rectangle is \(9\text{ m}\) long and \(4\text{ m}\) wide. Find its perimeter.
Show Answer
Step-by-step solution: \(2(9+4)=\boxed{26\text{ m}}\).
3Question 3Intermediate
An L-shaped path has six exposed edges measuring \(8,3,2,4,6,7\text{ m}\). Find its perimeter.
Show Answer
Step-by-step solution: Add all six lengths: \(8+3+2+4+6+7=\boxed{30\text{ m}}\).
4Question 4Exam-style
An edging strip surrounds a \(2.4\text{ m}\) by \(75\text{ cm}\) tabletop. How many metres of strip are needed?
Show Answer
Step-by-step solution: Convert \(75\text{ cm}=0.75\text{ m}\). \(2(2.4+0.75)=\boxed{6.3\text{ m}}\).
The area of a rectangle
5Question 5Foundation
Find the area of a \(9\text{ cm}\) by \(5\text{ cm}\) rectangle.
Show Answer
Step-by-step solution: \(A=9\times5=\boxed{45\text{ cm}^2}\).
6Question 6Foundation
A square has side \(11\text{ m}\). Find its area.
Show Answer
Step-by-step solution: \(11^2=\boxed{121\text{ m}^2}\).
7Question 7Intermediate
A rectangle has area \(96\text{ m}^2\) and length \(12\text{ m}\). Find its width.
Show Answer
Step-by-step solution: Width \(=96/12=\boxed{8\text{ m}}\).
8Question 8Exam-style
A garden is \(250\text{ cm}\) by \(4\text{ m}\). Find its area in \(\text{m}^2\).
Show Answer
Step-by-step solution: Convert \(250\text{ cm}=2.5\text{ m}\); area \(=2.5\times4=\boxed{10\text{ m}^2}\).
The area of a triangle
9Question 9Foundation
Find the area of a triangle with base \(10\text{ cm}\) and height \(6\text{ cm}\).
Show Answer
Step-by-step solution: \(\tfrac12(10)(6)=\boxed{30\text{ cm}^2}\).
10Question 10Foundation
A triangle has base \(7\text{ m}\) and height \(4\text{ m}\). Find its area.
Show Answer
Step-by-step solution: \(\tfrac12(7)(4)=\boxed{14\text{ m}^2}\).
11Question 11Intermediate
Find the base of a triangle with area \(45\text{ cm}^2\) and perpendicular height \(9\text{ cm}\).
Show Answer
Step-by-step solution: \(45=\tfrac12 b(9)\), so \(b=90/9=\boxed{10\text{ cm}}\).
12Question 12Exam-style
A triangular fabric panel has base \(1.8\text{ m}\) and height \(95\text{ cm}\). Find its area in \(\text{m}^2\).
Show Answer
Step-by-step solution: \(95\text{ cm}=0.95\text{ m}\); \(A=\tfrac12(1.8)(0.95)=\boxed{0.855\text{ m}^2}\).
The area of a parallelogram
13Question 13Foundation
A parallelogram has base 11 cm and perpendicular height 3 cm. Find its area.
Show Answer
Step-by-step solution: \(11\times3=\boxed{33\text{ cm}^2}\).
14Question 14Foundation
The area of a parallelogram is \(72\text{ cm}^2\) and its base is 8 cm. Find the height.
Show Answer
Step-by-step solution: \(h=72/8=\boxed{9\text{ cm}}\).
15Question 15Intermediate
A parallelogram has base 4.5 m and height 2.2 m. Find its area.
Show Answer
Step-by-step solution: \(4.5\times2.2=\boxed{9.9\text{ m}^2}\).
16Question 16Exam-style
A field is a parallelogram with base 32 m, height 18 m and side length 25 m. Find the area and perimeter.
Show Answer
Step-by-step solution: Area \(=32\times18=\boxed{576\text{ m}^2}\). Perimeter \(=2(32+25)=\boxed{114\text{ m}}\).
The area of a trapezium
17Question 17Foundation
Find the area of a trapezium with parallel sides 6 and 10 cm and height 4 cm.
Show Answer
Step-by-step solution: \(\tfrac12(6+10)\times4=\boxed{32\text{ cm}^2}\).
18Question 18Foundation
The parallel sides are 9 m and 15 m with height 3 m. Find the area.
Show Answer
Step-by-step solution: \(\tfrac12(9+15)3=\boxed{36\text{ m}^2}\).
19Question 19Intermediate
Area 65 cm²; parallel sides 8 and 18 cm. Find height.
Show Answer
Step-by-step solution: \(65=\tfrac12(26)h=13h\), so \(h=\boxed{5\text{ cm}}\).
20Question 20Exam-style
A trapezium has area 110 cm², height 10 cm, and one parallel side 8 cm. Find the other.
Show Answer
Step-by-step solution: \(110=5(8+b)\); \(22=8+b\), hence \(b=\boxed{14\text{ cm}}\).
The area of shapes made from rectangles and triangles
21Question 21Foundation
A 7 cm by 5 cm rectangle has a 2 cm by 3 cm notch removed. Find the area.
Show Answer
Step-by-step solution: \(7(5)-2(3)=35-6=\boxed{29\text{ cm}^2}\).
22Question 22Foundation
A shape is made from non-overlapping rectangles of area 24 and 18 cm². Find total area.
Show Answer
Step-by-step solution: Add: \(24+18=\boxed{42\text{ cm}^2}\).
23Question 23Intermediate
A triangle of base 8 cm and height 6 cm sits on a 10 by 6 cm rectangle. Find total area.
Show Answer
Step-by-step solution: \(10(6)+\tfrac12(8)(6)=60+24=\boxed{84\text{ cm}^2}\).
24Question 24Exam-style
From a 20 m by 12 m garden, remove two triangular flower beds, each of base 5 m and height 4 m. Find the remaining area.
Show Answer
Step-by-step solution: Whole area \(240\); two triangles each \(10\), so \(240-2(10)=\boxed{220\text{ m}^2}\).
The circumference of a circle
25Question 25Foundation
Find the circumference of a circle of radius 5 cm in terms of \(\pi\).
Show Answer
Step-by-step solution: \(2\pi(5)=\boxed{10\pi\text{ cm}}\).
26Question 26Foundation
Find the circumference of a circle with diameter 12 cm, correct to 1 d.p.
Show Answer
Step-by-step solution: \(C=12\pi=\boxed{37.7\text{ cm}}\).
27Question 27Intermediate
A circle has circumference \(18\pi\) cm. Find its diameter.
Show Answer
Step-by-step solution: \(d=C/\pi=\boxed{18\text{ cm}}\).
28Question 28Exam-style
A circular pond has radius 3 m. An edging strip surrounds its boundary. How long must the strip be to 1 d.p.?
Show Answer
Step-by-step solution: \(2\pi(3)=6\pi=\boxed{18.8\text{ m}}\).
The area of a circle
29Question 29Foundation
Find the area of a circle with radius 4 cm in terms of \(\pi\).
Show Answer
Step-by-step solution: \(\pi 4^2=\boxed{16\pi\text{ cm}^2}\).
30Question 30Foundation
Diameter 10 m. Find circle area in terms of \(\pi\).
Show Answer
Step-by-step solution: Radius 5 m, area \(=\boxed{25\pi\text{ m}^2}\).
31Question 31Intermediate
A circle has area \(49\pi\text{ cm}^2\). Find the radius.
Show Answer
Step-by-step solution: \(r^2=49\), so \(r=\boxed{7\text{ cm}}\).
32Question 32Exam-style
A circular lawn has radius 8 m. A circular flowerbed of radius 2 m is removed. Find the remaining area in terms of \(\pi\).
Show Answer
Step-by-step solution: \(\pi(8^2-2^2)=\pi(64-4)=\boxed{60\pi\text{ m}^2}\).
Arc length and sector area
33Question 33Foundation
Find the arc length of a semicircle of radius 5 cm in terms of \(\pi\).
Show Answer
Step-by-step solution: \((180/360)2\pi(5)=\boxed{5\pi\text{ cm}}\).
34Question 34Foundation
A sector has angle 120° and radius 3 cm. Find its area in terms of \(\pi\).
Show Answer
Step-by-step solution: \((120/360)\pi(3^2)=\boxed{3\pi\text{ cm}^2}\).
35Question 35Intermediate
A 45° sector of radius 8 cm has what arc length?
Show Answer
Step-by-step solution: \((45/360)2\pi(8)=\boxed{2\pi\text{ cm}}\).
36Question 36Exam-style
A major sector has angle 300° and radius 6 cm. Find its area and perimeter.
Show Answer
Step-by-step solution: Area \(=(300/360)36\pi=\boxed{30\pi\text{ cm}^2}\). Arc \(=(300/360)12\pi=10\pi\), perimeter \(=12+10\pi\approx\boxed{43.4\text{ cm}}\).
The volume of a prism
37Question 37Foundation
Find the volume of a cuboid 5 cm by 3 cm by 2 cm.
Show Answer
Step-by-step solution: \(5(3)(2)=\boxed{30\text{ cm}^3}\).
38Question 38Foundation
A prism has cross-sectional area 18 cm² and length 9 cm. Find volume.
Show Answer
Step-by-step solution: \(18(9)=\boxed{162\text{ cm}^3}\).
39Question 39Intermediate
Find the exact volume of a cylinder with radius 3 cm and height 7 cm.
Show Answer
Step-by-step solution: \(\pi(3)^2(7)=\boxed{63\pi\text{ cm}^3}\).
40Question 40Exam-style
A triangular prism has a 10 cm base, 6 cm perpendicular cross-sectional height and length 2 m. Find volume in cm³.
Show Answer
Step-by-step solution: Convert 2 m = 200 cm; cross-section \(=30\text{ cm}^2\); \(V=30(200)=\boxed{6000\text{ cm}^3}\).
The surface area of a prism
41Question 41Foundation
A cube has edge 4 cm. Find total surface area.
Show Answer
Step-by-step solution: 6 square faces each \(4^2=16\). \(S=6(16)=\boxed{96\text{ cm}^2}\).
42Question 42Foundation
A closed cylinder has radius 2 cm and height 6 cm. Find total area in terms of \(\pi\).
Show Answer
Step-by-step solution: \(2\pi(2)(6)+2\pi(2^2)=24\pi+8\pi=\boxed{32\pi\text{ cm}^2}\).
43Question 43Intermediate
A triangular prism has cross-sectional area 12 cm², perimeter 16 cm and length 8 cm. Find its surface area.
Show Answer
Step-by-step solution: \(S=2(12)+16(8)=24+128=\boxed{152\text{ cm}^2}\).
44Question 44Exam-style
An open box has a 10 cm by 8 cm base and height 6 cm. Find material area, ignoring thickness.
Show Answer
Step-by-step solution: Base \(80\); two faces \(10(6)\) and two \(8(6)\): \(80+120+96=\boxed{296\text{ cm}^2}\).
The volume of a pyramid, a cone and a sphere
45Question 45Foundation
Find the volume of a cone with radius 2 cm and height 9 cm.
Show Answer
Step-by-step solution: \(\tfrac13\pi(4)(9)=\boxed{12\pi\text{ cm}^3}\).
46Question 46Foundation
A square pyramid has base side 4 cm and height 12 cm. Find its volume.
Show Answer
Step-by-step solution: \(\tfrac13(4^2)(12)=\boxed{64\text{ cm}^3}\).
47Question 47Intermediate
Find the volume of a sphere of radius 3 cm in terms of \(\pi\).
Show Answer
Step-by-step solution: \(\tfrac43\pi(27)=\boxed{36\pi\text{ cm}^3}\).
48Question 48Exam-style
A cone of height 15 cm has volume \(125\pi\text{ cm}^3\). Find its base radius.
Show Answer
Step-by-step solution: \(125\pi=\tfrac13\pi r^2(15)=5\pi r^2\). Hence \(r^2=25\) and \(r=\boxed{5\text{ cm}}\).
The surface area of a pyramid, a cone and a sphere
49Question 49Foundation
Find the surface area of a sphere with radius 2 cm.
Show Answer
Step-by-step solution: \(4\pi(2^2)=\boxed{16\pi\text{ cm}^2}\).
50Question 50Foundation
A cone has radius 3 cm and slant height 10 cm. Find its curved area.
Show Answer
Step-by-step solution: \(\pi(3)(10)=\boxed{30\pi\text{ cm}^2}\).
51Question 51Intermediate
A closed cone has radius 3 cm and slant height 5 cm. Find its total surface area.
Show Answer
Step-by-step solution: \(\pi(3)(5)+\pi(3)^2=15\pi+9\pi=\boxed{24\pi\text{ cm}^2}\).
52Question 52Exam-style
A cone has height 15 cm and radius 8 cm. Find its curved surface area.
Show Answer
Step-by-step solution: Slant height \(=\sqrt{8^2+15^2}=17\). Curved area \(=\boxed{136\pi\text{ cm}^2}\).
The area and volume of compound shapes
53Question 53Foundation
A 200 cm³ prism has a 45 cm³ piece removed. Find the remaining volume.
Show Answer
Step-by-step solution: \(200-45=\boxed{155\text{ cm}^3}\).
54Question 54Foundation
A hemisphere of radius 3 cm sits on a cylinder of radius 3 cm and height 6 cm. Find total volume in terms of \(\pi\).
Show Answer
Step-by-step solution: Cylinder \(=\pi(9)(6)=54\pi\). Hemisphere \(=\tfrac12\times\tfrac43\pi(27)=18\pi\). Total \(=\boxed{72\pi\text{ cm}^3}\).
55Question 55Intermediate
A 60° minor sector in a circle of radius 6 cm forms a segment. Find segment area in exact form.
Show Answer
Step-by-step solution: Sector \(=(60/360)36\pi=6\pi\). Triangle \(=\tfrac12(6)(6)\sin60^\circ=9\sqrt3\). Segment \(=\boxed{6\pi-9\sqrt3\text{ cm}^2}\).
56Question 56Exam-style
A cone of height 15 cm, radius 9 cm has its top 5 cm cut off parallel to the base. Find remaining volume.
Show Answer
Step-by-step solution: By similarity the small radius is \(9(5/15)=3\text{ cm}\). Subtract: \(\tfrac13\pi(9^2)(15)-\tfrac13\pi(3^2)(5)=405\pi-15\pi=\boxed{390\pi\text{ cm}^3}\).
57Question 57Exam-style challenge
A solid metal frustum has a vertical height of \(4\text{ cm}\), a lower radius of \(9\text{ cm}\) and an upper radius of \(6\text{ cm}\). Both circular ends are closed. Calculate its total external surface area, giving your answer in terms of \(\pi\).
Show Answer
Step 1 — Find the slant height: The difference between the radii is \(9-6=3\text{ cm}\). Using Pythagoras, \(s=\sqrt{4^2+3^2}=5\text{ cm}\).
Step 2 — Find the curved surface area: \(\pi(R+r)s=\pi(9+6)(5)=75\pi\text{ cm}^2\).
Step 3 — Add both circular ends: \(\pi(9^2)+\pi(6^2)=117\pi\text{ cm}^2\).
Total external surface area: \(75\pi+117\pi=\boxed{192\pi\text{ cm}^2}\).
High-grade revision checklist
- Can I solve unfamiliar questions using The perimeter of a 2-D shape and explain my method?
- Can I solve unfamiliar questions using The area of a rectangle and explain my method?
- Can I solve unfamiliar questions using The area of a triangle and explain my method?
- Can I solve unfamiliar questions using The area of a parallelogram and explain my method?
- Can I solve unfamiliar questions using The area of a trapezium and explain my method?
- Can I solve unfamiliar questions using The area of shapes made from rectangles and triangles and explain my method?
- Can I solve unfamiliar questions using The circumference of a circle and explain my method?
- Can I solve unfamiliar questions using The area of a circle and explain my method?
- Can I solve unfamiliar questions using Arc length and sector area and explain my method?
- Can I solve unfamiliar questions using The volume of a prism and explain my method?
- Can I solve unfamiliar questions using The surface area of a prism and explain my method?
- Can I solve unfamiliar questions using The volume of a pyramid, a cone and a sphere and explain my method?
- Can I solve unfamiliar questions using The surface area of a pyramid, a cone and a sphere and explain my method?
- Can I solve unfamiliar questions using The area and volume of compound shapes and explain my method?
Before the exam, practise without looking at the answers, check units and rounding, and write geometric or statistical reasons where they are needed.