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KS3 Maths · Year 9Chapter 9 · Similarity and circle theorems
Year 9 · Chapter 9

Similarity and circle theorems

By the end of this chapter you will be able to

  • Use the scale factor for lengths, areas and volumes of similar shapes
  • Find the scale factor from a known pair of lengths, areas or volumes
  • Find missing lengths in similar triangles, and in triangles formed by parallel lines
  • State and use circle theorems, giving a reason for each step
  • Write a short geometric argument that uses circle theorems, with a reason for every step

You should already know

  • Enlargement and scale factors for lengths (Year 8, Chapter 11)
  • Angles in a triangle and angles on a straight line (Year 8, Chapter 9)
  • Parts of a circle: centre, radius, diameter, chord, tangent (Year 8, Chapter 10)

Where this leads

  • Pythagoras and trigonometry in right-angled triangles (Year 9, Chapter 8)
  • Vectors and compound measures (Year 9, Chapter 10)
  • Circle theorems proofs and the alternate segment theorem in GCSE

National Curriculum for England, Key Stage 3 mathematics, Year 9. Content: Use scale factors for areas (k2) and volumes (k3) of similar shapes, Find missing lengths in similar triangles, including triangles formed by a line parallel to one side, Use circle theorems with reasons: angle in a semicircle, tangent perpendicular to the radius, angle at the centre twice the angle at the circumference, angles in the same segment, opposite angles of a cyclic quadrilateral, and isosceles triangles from radii, Give a reason for each step in a multi-step geometric argument involving circles.

Exercise P  Do you remember?

Non-calculator
1
FWhat is the scale factor if a length of 4 cm becomes 12 cm?
2
FIn a triangle with angles 50° and 60°, what is the third angle?
3
FWhat is the area of a square with side 5 cm?
4
FWhich two sides are equal in an isosceles triangle?
5
FA line from the centre of a circle to its edge is called what?
9.1

Scale factors for lengths, areas and volumes

Year 9
Use scale factors for areas (k2) and volumes (k3) of similar shapes
Video lesson: Scale factors for lengths, areas and volumesComing soon
A model of a sports car is made with a scale factor of 1 : 10. The length of the model is one tenth of the car. Its area of the bonnet is one hundredth, and its volume is one thousandth, of the real car.

Two shapes are similar if one is an enlargement of the other. If the linear scale factor is k, then lengths are multiplied by k, areas by k2, and volumes by k3.

To find the scale factor from a pair of areas or volumes, take the square root (for areas) or the cube root (for volumes).

Key ideas

  • Lengths scale by k.
  • Areas scale by k2.
  • Volumes scale by k3.
  • To find k from an area ratio, take the square root of the ratio.
Worked example 9.1.1
Two similar shapes have a scale factor of 4. A length of 3 m on the small shape. Find the length on the large shape.

Method. Multiply each length by the scale factor.

  1. Identify the scale factor. The lengths are multiplied by the scale factor k = 4.
    k = 4
  2. Multiply the length. New length = k × old length.
    4 × 3 = 12
  3. Write the answer with the unit.
    12 m

Check. Dividing 12 by 4 gives back 3.

Watch out. Lengths scale by k, areas by k2, and volumes by k3.

Worked example 9.1.2
A small shape has area 12 cm2 and the scale factor is 3. Find the area of the large shape.

Method. Square the linear scale factor to get the area scale factor.

  1. Find the area scale factor. Areas scale by k2, so the area scale factor is 32 = 9.
    32 = 9
  2. Multiply the area. New area = area scale factor × old area.
    9 × 12 = 108
  3. Write the answer with the unit.
    108 cm2

Check. 12 × 9 = 108.

Watch out. Do not multiply the area by k. Areas scale by k2.

Worked example 9.1.3
A small solid has volume 40 cm3. The scale factor is 2. Find the volume of the large solid.

Method. Cube the linear scale factor to get the volume scale factor.

  1. Find the volume scale factor. Volumes scale by k3, so the volume scale factor is 23 = 8.
    23 = 8
  2. Multiply the volume. New volume = volume scale factor × old volume.
    8 × 40 = 320
  3. Write the answer with the unit.
    320 cm3

Check. 40 × 8 = 320.

Watch out. A volume scales by k3. Doubling the lengths multiplies the volume by 8.

Watch out

Watch out: doubling the lengths of a shape does not double its area. The area is multiplied by 22 = 4.

Exam tip

Exam tip. Write the scale factor for lengths first, then square or cube it for area or volume.

Exercise 9A  Scale factors for lengths, areas and volumes

Non-calculator
1
FA shape is enlarged with scale factor 3. A length of 4 cm becomes what length?
2
FA shape is enlarged with scale factor 5. A length of 6 m becomes what length?
3
FA similar shape has linear scale factor 2. Its area is 15 cm2. Find the area of the new shape.
4
FA similar shape has linear scale factor 3. Its area is 12 cm2. Find the area of the new shape.
5
FA similar solid has linear scale factor 2. Its volume is 40 cm3. Find the volume of the new solid.
6
FA similar solid has linear scale factor 3. Its volume is 5 cm3. Find the volume of the new solid.
7
FTwo similar rectangles have areas 12 cm2 and 108 cm2. Find the linear scale factor from the small to the large rectangle.
8
FTwo similar shapes have areas 9 cm2 and 36 cm2. Find the linear scale factor from the small to the large shape.
9
PSA model boat is built at scale 1 : 50. The area of a sail on the model is 4 m2. Find the area of the real sail.
10
PSTwo similar bottles have heights in the ratio 2 : 3. The small bottle holds 64 ml. How much does the large bottle hold?
11
PSTwo similar cuboids have lengths in the ratio 2 : 5. The small one has volume 40 cm3. Find the volume of the large one.
12
RExplain why doubling every length of a cube multiplies its volume by 8, not by 2.
9.2

Similar triangles and lines parallel to a side

Year 9
Find missing lengths in similar triangles, including parallel lines
Video lesson: Similar triangles and lines parallel to a sideComing soon
A surveyor measures a distance across a river using two similar triangles. A stick 2 m tall casts a shadow of 3 m, and a tree casts a shadow of 12 m. The triangles are similar, so the heights are in the same ratio as the shadows.

Two triangles are similar if their angles are equal and their sides are in the same ratio. Find the scale factor from one pair of matching sides, then use it to find the other missing sides.

If a line is drawn parallel to one side of a triangle, it cuts off a smaller triangle that is similar to the original.

Key ideas

  • Matching sides are the sides in the same position in each triangle.
  • Scale factor = new side ÷ old side (for the matching pair).
  • A line parallel to one side of a triangle creates a similar triangle.
  • Use the scale factor from one pair to find the missing side.
Worked example 9.2.1
Two similar triangles have matching sides 5 cm and 15 cm. A side of 4 cm in the small triangle matches a side in the large triangle. Find it.

Method. Use the scale factor from the pair of matching sides.

  1. Find the scale factor. Divide a big side by the matching small side.
    k = 15 ÷ 5 = 3
  2. Find the missing side. Multiply the matching small side by the scale factor.
    4 × 3 = 12
  3. Write the answer.
    12

Check. 12 ÷ 4 = 3, the same scale factor.

Watch out. Match the sides carefully. The longest side in one triangle matches the longest side in the other.

Worked example 9.2.2
Two similar triangles have matching sides 6 cm and 9 cm. The side 8 cm in the small triangle matches a side in the large one. Find that side.

Method. Use the scale factor from the pair of matching sides.

  1. Find the scale factor. Divide a big side by the matching small side.
    k = 9 ÷ 6 = 3/2
  2. Find the missing side. Multiply the matching small side by the scale factor.
    8 × 3/2 = 12
  3. Write the answer.
    12

Check. 12 ÷ 8 = 3/2, the same scale factor.

Watch out. Match the sides carefully. The longest side in one triangle matches the longest side in the other.

Worked example 9.2.3
In triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 4 cm, DB = 6 cm and DE = 5 cm. Find BC.

Method. Parallel lines give similar triangles, so the sides are in the same ratio.

  1. Find the scale factor. Triangle ADE is similar to triangle ABC, because DE is parallel to BC.
    AB = AD + DB = 4 + 6 = 10, so the scale factor is 10 ÷ 4 = 5/2
  2. Find BC. Multiply DE by the scale factor.
    BC = 5 × 5/2 = 12.5
  3. Write the answer.
    12.5

Check. DE / BC = 5 ÷ 12.5 = 2/5 = AD / AB, as required.

Watch out. The scale factor is AB ÷ AD, not DB ÷ AD.

Watch out

Watch out: a scale factor is the ratio of a matching pair. Do not divide sides from different triangles.

Exam tip

Exam tip. Write the matching pairs as fractions, such as 6/9, before you find the scale factor.

Exercise 9B  Similar triangles and parallel lines

Non-calculator
1
FTwo similar triangles have matching sides 6 cm and 9 cm. A side 8 cm in the small triangle matches a side in the large triangle. Find that side.
2
FTwo similar triangles have matching sides 4 cm and 10 cm. A side 6 cm in the small triangle matches a side in the large triangle. Find that side.
3
FTwo similar triangles have matching sides 5 cm and 8 cm. A side 7 cm in the small triangle matches a side in the large triangle. Find that side.
4
FTwo similar triangles have matching sides 3 cm and 12 cm. A side 5 cm in the small triangle matches a side in the large triangle. Find that side.
5
FTwo similar triangles have matching sides 10 cm and 15 cm. A side 8 cm in the small triangle matches a side in the large triangle. Find that side.
6
FTwo similar triangles have matching sides 2 cm and 7 cm. A side 3 cm in the small triangle matches a side in the large triangle. Find that side.
7
PSIn triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 4 cm, DB = 6 cm and DE = 5 cm. Find BC.
DEADAEBCABAC
8
PSIn triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 3 cm, DB = 9 cm and DE = 4 cm. Find BC.
DEADAEBCABAC
9
PSIn triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 5 cm, DB = 10 cm and DE = 6 cm. Find BC.
DEADAEBCABAC
10
PSIn triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 2 cm, DB = 6 cm and DE = 3 cm. Find BC.
DEADAEBCABAC
11
PSA stick 2 m tall casts a shadow 3 m long. A tree casts a shadow 12 m long at the same time. Find the height of the tree.
12
RExplain why triangle ADE is similar to triangle ABC when DE is parallel to BC.
9.3

Circle theorems: semicircles, tangents and radii

Year 9
Use circle theorems with reasons: semicircle, tangent perpendicular to the radius, isosceles triangles from radii
Video lesson: Circle theorems: semicircles, tangents and radiiComing soon
A circular table has a diameter AB. A diner sits at a point C on the edge of the table. Whatever point C is chosen, the angle ACB is the same. This is a circle theorem, and it can be explained with a reason.

A circle theorem is a fact about circles that is always true. Each step in an argument needs a reason, such as “angle in a semicircle”, “tangent perpendicular to radius” or “base angles of an isosceles triangle”.

Key ideas

  • Angle in a semicircle is 90°: if AB is a diameter and C is on the circle, angle ACB = 90°.
  • A tangent is perpendicular to the radius at the point of contact: the angle is 90°.
  • Two radii make an isosceles triangle: the base angles are equal.
  • Always give the reason with each step.
Worked example 9.3.1
AB is a diameter of a circle with centre O, and C is a point on the circle. Explain why angle ACB is 90°.

Method. Use the theorem: the angle in a semicircle.

  1. Name the theorem. The angle in a semicircle is a right angle.
    angle in a semicircle
  2. Apply it. AB is a diameter, so the triangle ACB is in a semicircle.
    angle ACB
  3. Write the answer. The reason is the angle in a semicircle.
    angle ACB = 90°

Check. Any triangle with its longest side as a diameter has a right angle opposite the diameter.

Worked example 9.3.2
OA = OB are radii of a circle and angle AOB = 40°. Find angle OAB and give a reason.

Method. Use the isosceles triangle formed by two radii.

  1. Find the other angles. Two radii are equal, so triangle OAB is isosceles with OA = OB.
    angle OAB = angle OBA
  2. Use the angle sum. The angles of a triangle add up to 180°.
    180 − 40 = 140
  3. Share equally. The two base angles are equal.
    140 ÷ 2 = 70
  4. Write the answer. Reason: base angles of an isosceles triangle are equal.
    angle OAB = 70°

Check. 70 + 70 + 40 = 180, as the angle sum requires.

Worked example 9.3.3
A tangent touches a circle at T, and O is the centre. P is a point on the tangent. Explain why angle OTP is 90°.

Method. Use the tangent and radius theorem.

  1. Name the radius. OT is a radius, drawn to the point of contact T.
    OT is a radius
  2. Name the theorem. A tangent is perpendicular to the radius at the point of contact.
    tangent ⊥ radius
  3. Write the answer. Reason: tangent perpendicular to radius.
    angle OTP = 90°

Check. The tangent is at a right angle to the radius at T.

Watch out

Watch out: a tangent is always at 90° to the radius at the point of contact, not at 90° to any other line.

Exam tip

Exam tip. Write the name of the theorem next to each angle you find. It is the reason you need in the exam.

Exercise 9C  Semicircles, tangents and radii

Non-calculator
1
FAB is a diameter of a circle and C is any other point on the circle. Find angle ACB, and give a reason.
OABC
2
FA tangent touches a circle at P, and O is the centre. Find the angle between the tangent and the radius OP, and give a reason.
3
FOA = OB are radii of a circle and angle AOB = 40°. Find angle OAB, giving a reason.
OAB
4
FOA = OB are radii and angle OAB = 55°. Find angle AOB, giving a reason.
5
PSA tangent meets the circle at T, and the radius OT is drawn. The angle between OT and the tangent is 90°. Angle TOP is 35° where P is on the tangent. Find angle OPT and give a reason.
6
PSAB is a diameter and C is a point on the circle with angle CAB = 32°. Find angle ABC, giving a reason.
7
ROA and OB are radii and the triangle OAB has angles 22.5° at A and B. Find angle AOB, and explain why the triangle is isosceles.
9.4

Angle at the centre, the same segment and cyclic quadrilaterals

Year 9
Use the angle at the centre, angles in the same segment and cyclic quadrilaterals
Video lesson: Angle at the centre, the same segment and cyclic quadrilateralsComing soon
A Ferris wheel has a centre O. Two seats A and B are on the wheel. The angle AOB at the centre is twice the angle that the same arc makes at a seat C on the rim. Circle theorems link angles at the centre and at the edge.

The angle at the centre is twice the angle at the circumference standing on the same arc. Angles in the same segment (on the same arc) are equal. In a cyclic quadrilateral (all four corners on the circle), opposite angles add up to 180°.

Key ideas

  • Angle at the centre = 2 × angle at the circumference (same arc).
  • Angles in the same segment (on the same arc) are equal.
  • Opposite angles of a cyclic quadrilateral add up to 180°.
  • Give the theorem as the reason for each step.
Worked example 9.4.1
Angle AOB is 124° at the centre O. Find the angle ACB at a point C on the circle, on the same arc.

Method. Use the angle at the centre theorem.

  1. State the theorem. The angle at the centre is twice the angle at the circumference on the same arc.
    angle AOB = 2 × angle ACB
  2. Rearrange. Divide the angle at the centre by 2.
    angle ACB = 124 ÷ 2
  3. Write the answer. Reason: angle at the centre is twice the angle at the circumference.
    angle ACB = 62°

Check. 2 × 62 = 124.

Worked example 9.4.2
Angle ACB is 35°. Find angle ADB, where D is another point on the circle on the same arc AB.

Method. Use the same segment theorem.

  1. Check the arc. Both angles stand on the same arc AB, with C and D on the same side.
    arc AB
  2. State the theorem. Angles in the same segment are equal.
    angle ACB = angle ADB
  3. Write the answer. Reason: angles in the same segment are equal.
    angle ADB = 35°

Check. Both angles stand on the same arc AB.

Worked example 9.4.3
ABCD is a cyclic quadrilateral and angle ABC = 72°. Find angle ADC, and give a reason.

Method. Use the opposite angles of a cyclic quadrilateral.

  1. State the theorem. Opposite angles of a cyclic quadrilateral add up to 180°.
    angle ABC + angle ADC = 180
  2. Work out the angle. 180 − 72 = 108.
    180 − 72 = 108
  3. Write the answer. Reason: opposite angles of a cyclic quadrilateral.
    angle ADC = 108°

Check. 72 + 108 = 180.

Watch out

Watch out: the angle at the centre is twice the angle at the circumference only when both stand on the same arc.

Exam tip

Exam tip. Draw the arc on your diagram, so you can see which angles stand on it.

Exercise 9D  Angle at the centre, same segment and cyclic quadrilaterals

Non-calculator
1
FAngle AOB is 124° at the centre O. Find the angle ACB at a point C on the circle, on the same arc. Give a reason.
OABC
2
FAngle AOB is 90° at the centre O. Find the angle ACB at a point C on the circle, on the same arc. Give a reason.
OABC
3
FAngle AOB is 150° at the centre O. Find the angle ACB at a point C on the circle, on the same arc. Give a reason.
OABC
4
FAngle AOB is 80° at the centre O. Find the angle ACB at a point C on the circle, on the same arc. Give a reason.
OABC
5
FAngle AOB is 110° at the centre O. Find the angle ACB at a point C on the circle, on the same arc. Give a reason.
OABC
6
FAngle ACB is 35°. Find angle ADB, where D is another point on the circle on the same arc AB. Give a reason.
OABCD
7
FAngle ACB is 48°. Find angle ADB, where D is another point on the circle on the same arc AB. Give a reason.
OABCD
8
FAngle ACB is 62°. Find angle ADB, where D is another point on the circle on the same arc AB. Give a reason.
OABCD
9
FAngle ACB is 27°. Find angle ADB, where D is another point on the circle on the same arc AB. Give a reason.
OABCD
10
PSABCD is a cyclic quadrilateral and angle ABC = 72°. Find angle ADC, and give a reason.
11
PSABCD is a cyclic quadrilateral and angle ABC = 85°. Find angle ADC, and give a reason.
12
PSABCD is a cyclic quadrilateral and angle ABC = 110°. Find angle ADC, and give a reason.
13
PSABCD is a cyclic quadrilateral and angle ABC = 64°. Find angle ADC, and give a reason.
14
PSO is the centre of a circle, and angle AOB = 130°. C is a point on the major arc. Find angle ACB, giving a reason.
9.5

Giving reasons in a geometric argument

Year 9
Give a reason for each step in a multi-step geometric argument involving circles
Video lesson: Giving reasons in a geometric argumentComing soon
A student is asked to prove that a triangle with a diameter as one side has a right angle. Each line of the argument must be supported by a reason, so that a reader can check it.

A geometric argument is a list of steps, each with a reason. Reasons come from facts you know: angle sum of a triangle, isosceles triangle, circle theorem, or parallel lines.

Key ideas

  • Write one step per line, and give a reason for each step.
  • Name the theorem in the reason, such as “angle in a semicircle”.
  • Check that every given fact is used.
  • A final line states what you have shown.
Worked example 9.5.1
Triangle OAB has OA = OB and angle AOB = 60°. Show that triangle OAB is equilateral.

Method. Use isosceles triangles and the angle sum.

  1. Radii are equal. OA and OB are radii, so OA = OB.
    OA = OB
  2. Base angles. The base angles of the isosceles triangle are equal.
    angle OAB = angle OBA
  3. Angle sum. The angles add up to 180°: 180 − 60 = 120, and 120 ÷ 2 = 60.
    angle OAB = 60
  4. Write the conclusion. All three angles are 60°, so the triangle is equilateral.
    Shown

Check. Three equal angles of 60° make an equilateral triangle.

Worked example 9.5.2
OA and OB are radii of a circle. Show that angle OAB = angle OBA.

Method. Use the fact that two radii are equal.

  1. Radii are equal. OA and OB are both radii of the same circle.
    OA = OB
  2. Isosceles triangle. A triangle with two equal sides is isosceles.
    triangle OAB is isosceles
  3. Base angles. The angles opposite the equal sides are equal.
    angle OAB = angle OBA
  4. Write the conclusion. Reason: base angles of an isosceles triangle are equal.
    Shown

Check. The equal sides are the two radii.

Worked example 9.5.3
Give reasons for each step to show that angle ACB = 90° when AB is a diameter.

Method. Use the angle in a semicircle theorem.

  1. Step 1. AB is a diameter, so O is the midpoint of AB.
    AO = OB = radius
  2. Step 2. OA, OB and OC are radii, so they are equal.
    OA = OC = OB
  3. Step 3. Triangles AOC and BOC are isosceles, so their base angles are equal.
    angle OAC = angle OCA, angle OBC = angle OCB
  4. Step 4. The angles of triangle ABC add up to 180°, and the two pairs of base angles make 90°.
    angle ACB = 90
  5. Conclusion. Reason: angle in a semicircle.
    angle ACB = 90°

Check. This shows the angle in a semicircle is 90°.

Watch out

Watch out: a reason such as “because it is a circle” is not enough. Name the theorem or property.

Exam tip

Exam tip. Write the reason in the same line as the step, so the marker can see the link.

Exercise 9E  Reasons in geometric arguments

Non-calculator
1
RGive a reason for the statement: triangle OAB is isosceles, where OA and OB are radii.
2
FGive the reason why the angle between a tangent and the radius at the point of contact is 90°.
3
FIn a cyclic quadrilateral ABCD, angle A = 80°. Find angle C and give a reason.
4
RExplain the reasons for each step: the angles in triangle ABC add up to 180° because they are the angles of a triangle.
5
FAB is a diameter of a circle. Give the reason why angle ACB is 90° for any point C on the circle.
6
FOA and OB are radii of a circle, and the base angle OAB of triangle OAB is 65°. Find angle AOB, giving a reason.
7
PSGive reasons for each step: angle AOB = 110°, so the angle at the circumference ACB = 55°.
8
RExplain why two angles in the same segment are equal.
9
RIn a circle, angle ACB = 35° and the angle subtended by the same arc at O is 70°. Explain why the angle at O is twice the angle at C.
10
RABCD is a quadrilateral with angle A = 100°, B = 80°, C = 80° and D = 100°. Explain why it is cyclic.

Chapter summary

SkillMethod
Lengths, areas, volumesScale by k, k2 and k3.
Scale factor from areasTake the square root of the area ratio. For volumes, take the cube root.
Similar trianglesFind the scale factor from matching sides, then use it for the missing side.
Parallel lineCreates a similar triangle. Use the ratio of the sides.
Circle theoremsSemicircle 90°, tangent ⊥ radius, isosceles from radii, angle at centre = 2 × circumference, same segment equal, cyclic opposite angles 180°.

Can you…?

  • I can use scale factors for lengths, areas and volumes
  • I can find a missing side in similar triangles, including parallel lines
  • I can use the circle theorems for semicircles, tangents and radii
  • I can use the angle at the centre, same segment and cyclic quadrilateral theorems
  • I can give a reason for each step in a geometric argument

Key terms

similar
shapes that have the same angles and sides in the same ratio
scale factor
the number that each length is multiplied by to get the matching length
tangent
a straight line that touches a circle at exactly one point
cyclic quadrilateral
a quadrilateral with all four corners on a circle
segment
the region between a chord and an arc of a circle

Exercise 9Z  Check-up

Non-calculator
1
Two similar shapes have linear scale factor 4. A length of 2.5 cm on the small shape. Find the length on the large shape.
2
A similar shape has linear scale factor 2. Its area is 7 cm2. Find the area of the new shape.
3
A similar solid has linear scale factor 2. Its volume is 9 cm3. Find the volume of the new solid.
4
Two similar triangles have matching sides 4 cm and 10 cm. A side 6 cm in the small triangle matches a side in the large triangle. Find that side.
5
In triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 3 cm, DB = 9 cm and DE = 4 cm. Find BC.
6
Two similar rectangles have areas 4 cm2 and 36 cm2. Find the linear scale factor.
7
AB is a diameter of a circle, and C is any point on the circle. Find angle ACB and give a reason.
8
OA and OB are radii and angle AOB = 50°. Find angle OAB, giving a reason.
9
Angle AOB is 146° at the centre. Find the angle ACB on the same arc, giving a reason.
10
ABCD is a cyclic quadrilateral with angle ABC = 118°. Find angle ADC, giving a reason.

Total 21 marks. Allow about 21 minutes.

Exercise 9X  Practice test questions

Non-calculator
1
FTwo similar shapes have linear scale factor 5. A length of 3 cm on the small shape. Find the length on the large shape.[1 mark]
2
FA similar shape has linear scale factor 3. Its area is 8 cm2. Find the area of the new shape.[2 marks]
3
FA similar solid has linear scale factor 3. Its volume is 2 cm3. Find the volume of the new solid.[2 marks]
4
FTwo similar shapes have areas 25 cm2 and 100 cm2. Find the linear scale factor from the small to the large shape.[2 marks]
5
PSIn triangle ABC, D is on AB and E is on AC so that DE is parallel to BC. AD = 5 cm, DB = 10 cm and DE = 6 cm. Find BC.[3 marks]
6
FTwo similar triangles have matching sides 8 cm and 12 cm. A side of 10 cm in the small triangle matches a side in the large triangle. Find that side.[2 marks]
7
RAB is a diameter of a circle with centre O, and C is a point on the circle. Explain why angle ACB is 90°. Give a reason for each step.[3 marks]
8
FO is the centre of a circle, and angle AOB = 96°. C is a point on the major arc. Find angle ACB, giving a reason.[2 marks]
9
FABCD is a cyclic quadrilateral. Angle ABC = 84°. Find angle ADC, giving a reason.[2 marks]
10
RA tangent touches a circle at T. OT is a radius. Explain why angle OTP is 90° when P is on the tangent.[2 marks]
← Chapter 8: Pythagoras and trigonometryChapter 10: Vectors and compound measures →

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