GCSE Maths: Chords (Higher)
Part of the GCSE Maths circle theorems chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Chords
A chord joins two points on a circle. There is a neat fact about the line from the centre that meets the chord at a right angle: the perpendicular from the centre to a chord bisects the chord. Bisects means cuts exactly in half. So if M is the point where the perpendicular from the centre O meets the chord AB, then AM = MB.
Reason to write: “the perpendicular from the centre to a chord bisects the chord”. The opposite is also true: a line from the centre to the midpoint of a chord is perpendicular to the chord. Why? Join O to A and B. These are radii, so OA = OB and triangle OAB is isosceles. The line from the apex O that meets the base at 90° splits the triangle into two identical right-angled triangles (they share OM, have equal hypotenuses OA = OB, and a right angle at M). So AM = MB.
Using it with Pythagoras. The right-angled triangle OMA has the radius as its hypotenuse, half the chord as one short side and the distance from the centre to the chord as the other short side: r2 = d2 + (half the chord)2. Given any two of the three lengths, you can find the third. Remember to halve the chord before you use Pythagoras, and to double at the end if you want the whole chord.
Using it with trigonometry. The perpendicular also halves the angle at the centre: triangle OAB is isosceles, so OM bisects angle AOB. In triangle OMA you can use sin, cos or tan with the half chord, the radius and the half angle (Chapter 20). The calculator exercise at the end of this section practises this.
Key ideas
- The perpendicular from the centre to a chord bisects the chord, and a line from the centre to the midpoint of a chord is perpendicular to it.
- Reason: “the perpendicular from the centre to a chord bisects the chord”.
- Right-angled triangle: hypotenuse = radius, short sides = half the chord and the distance from the centre.
- r2 = d2 + (half chord)2.
- The depth of a segment (the height of the cut-off part) = radius − d, when the segment is the smaller part.
Watch out
Use half the chord in Pythagoras. The most common mistake is to use the full chord as a side of the right-angled triangle.Key words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.
Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.
- Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.AM = 24 ÷ 2 = 12 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).OA2 = OM2 + AM2
132 = OM2 + 122 - Rearrange. Take the known square from both sides.OM2 = 169 − 144 = 25
- Square root. OM = √25 = 5 cm
Check. 52 + 122 = 25 + 144 = 169 = 132. ✓
Watch out. Use half the chord in the triangle, not the whole chord.
Method. The perpendicular from the centre bisects the chord. Find half the chord with Pythagoras, then double.
- Draw the triangle. The perpendicular from the centre meets the chord at M and makes a right-angled triangle OMA with OA as the hypotenuse.OA = 10 cm, OM = 6 cm
- Use Pythagoras to find AM. Subtract the square of the known short side from the square of the hypotenuse.AM2 = 102 − 62 = 100 − 36 = 64
AM = √64 = 8 cm - Double it to get the whole chord. The perpendicular from the centre bisects the chord, so AM is only half of it.AB = 2 × 8 = 16 cm (the perpendicular from the centre to a chord bisects the chord)
Check. 62 + 82 = 100, so OA = 10. ✓
Watch out. The triangle gives you half the chord. Remember to double.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle. The radius is its hypotenuse.
- Find half the chord. The perpendicular from the centre to a chord bisects the chord.AM = 30 ÷ 2 = 15 cm (the perpendicular from the centre to a chord bisects the chord)
- Use Pythagoras in triangle OMA. The angle at M is 90°, and the radius OA is the hypotenuse.OA2 = 82 + 152 = 64 + 225 = 289
- Square root. The radius is positive.OA = √289 = 17 cm
Check. The distance from the centre to B is the same: √(82 + 152) = 17. ✓
Watch out. Add the squares to get the hypotenuse.
Method. Use the perpendicular from the centre to the chord, then take the distance from the radius.
- Draw the picture. The surface of the water is a chord. The lowest point of the water is at the bottom of the circle, directly below the centre.Chord = 24 cm, radius = 13 cm
- Find the distance from the centre to the chord. The perpendicular from the centre bisects the chord, which gives a right-angled triangle with half the chord and the radius.Half the chord = 12 cm
Distance2 = 132 − 122 = 169 − 144 = 25
Distance = 5 cm - Work out the depth. The depth is the radius take away the distance from the centre to the chord.Depth = 13 − 5 = 8 cm
Check. Water is 8 cm deep and the centre is 5 cm above the surface: 8 + 5 = 13 cm, the radius. ✓
Watch out. The depth is not the distance from the centre to the chord. It is the radius minus that distance.
Method. Show triangles OAM and OBM are congruent, then the angles at M are equal and must each be 90°.
- Join OA and OB. These are radii, so they are equal.OA = OB (radii)
- Use the midpoint. The question says M is the midpoint of AB.AM = MB
- Shared side. OM is in both triangles.OM is common
- Congruent triangles. All three pairs of sides are equal (SSS).Triangles OAM and OBM are congruent (SSS).
- Match the angles at M. Congruent triangles have equal matching angles.angle OMA = angle OMB
- Use the straight line. A, M and B are in a straight line, so these two equal angles add up to 180°.each is 180° ÷ 2 = 90°. So OM is perpendicular to AB. ✓
Check. Test: radius 13, AM = MB = 12, OM = 5. Then 52 + 122 = 169 = 132, so the angle at M is 90°. ✓
Watch out. The proof shows the second half of the theorem: from the midpoint to the right angle.
Exercise 26H Chords (no calculator)
Non-calculatorKey words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.
Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.
- Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.AM = 6 ÷ 2 = 3 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).OA2 = OM2 + AM2
52 = OM2 + 32 - Rearrange. Take the known square from both sides.OM2 = 25 − 9 = 16
- Square root. OM = √16 = 4 cm
Check. 42 + 32 = 16 + 9 = 25 = 52. ✓
Watch out. Use half the chord in the triangle, not the whole chord.
Key words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.
Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.
- Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.AM = 16 ÷ 2 = 8 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).OA2 = OM2 + AM2
102 = OM2 + 82 - Rearrange. Take the known square from both sides.OM2 = 100 − 64 = 36
- Square root. OM = √36 = 6 cm
Check. 62 + 82 = 36 + 64 = 100 = 102. ✓
Watch out. Use half the chord in the triangle, not the whole chord.
Key words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.
Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.
- Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.AM = 30 ÷ 2 = 15 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).OA2 = OM2 + AM2
172 = OM2 + 152 - Rearrange. Take the known square from both sides.OM2 = 289 − 225 = 64
- Square root. OM = √64 = 8 cm
Check. 82 + 152 = 64 + 225 = 289 = 172. ✓
Watch out. Use half the chord in the triangle, not the whole chord.
Method. The perpendicular from the centre bisects the chord. Find half the chord with Pythagoras, then double.
- Draw the triangle. The perpendicular from the centre meets the chord at M and makes a right-angled triangle OMA with OA as the hypotenuse.OA = 5 cm, OM = 3 cm
- Use Pythagoras to find AM. Subtract the square of the known short side from the square of the hypotenuse.AM2 = 52 − 32 = 25 − 9 = 16
AM = √16 = 4 cm - Double it to get the whole chord. The perpendicular from the centre bisects the chord, so AM is only half of it.AB = 2 × 4 = 8 cm (the perpendicular from the centre to a chord bisects the chord)
Check. 32 + 42 = 25, so OA = 5. ✓
Watch out. The triangle gives you half the chord. Remember to double.
Method. The perpendicular from the centre bisects the chord. Find half the chord with Pythagoras, then double.
- Draw the triangle. The perpendicular from the centre meets the chord at M and makes a right-angled triangle OMA with OA as the hypotenuse.OA = 13 cm, OM = 12 cm
- Use Pythagoras to find AM. Subtract the square of the known short side from the square of the hypotenuse.AM2 = 132 − 122 = 169 − 144 = 25
AM = √25 = 5 cm - Double it to get the whole chord. The perpendicular from the centre bisects the chord, so AM is only half of it.AB = 2 × 5 = 10 cm (the perpendicular from the centre to a chord bisects the chord)
Check. 122 + 52 = 169, so OA = 13. ✓
Watch out. The triangle gives you half the chord. Remember to double.
Method. The perpendicular from the centre bisects the chord. Find half the chord with Pythagoras, then double.
- Draw the triangle. The perpendicular from the centre meets the chord at M and makes a right-angled triangle OMA with OA as the hypotenuse.OA = 10 cm, OM = 8 cm
- Use Pythagoras to find AM. Subtract the square of the known short side from the square of the hypotenuse.AM2 = 102 − 82 = 100 − 64 = 36
AM = √36 = 6 cm - Double it to get the whole chord. The perpendicular from the centre bisects the chord, so AM is only half of it.AB = 2 × 6 = 12 cm (the perpendicular from the centre to a chord bisects the chord)
Check. 82 + 62 = 100, so OA = 10. ✓
Watch out. The triangle gives you half the chord. Remember to double.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle. The radius is its hypotenuse.
- Find half the chord. The perpendicular from the centre to a chord bisects the chord.AM = 24 ÷ 2 = 12 cm (the perpendicular from the centre to a chord bisects the chord)
- Use Pythagoras in triangle OMA. The angle at M is 90°, and the radius OA is the hypotenuse.OA2 = 52 + 122 = 25 + 144 = 169
- Square root. The radius is positive.OA = √169 = 13 cm
Check. The distance from the centre to B is the same: √(52 + 122) = 13. ✓
Watch out. Add the squares to get the hypotenuse.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle. The radius is its hypotenuse.
- Find half the chord. The perpendicular from the centre to a chord bisects the chord.AM = 16 ÷ 2 = 8 cm (the perpendicular from the centre to a chord bisects the chord)
- Use Pythagoras in triangle OMA. The angle at M is 90°, and the radius OA is the hypotenuse.OA2 = 152 + 82 = 225 + 64 = 289
- Square root. The radius is positive.OA = √289 = 17 cm
Check. The distance from the centre to B is the same: √(152 + 82) = 17. ✓
Watch out. Add the squares to get the hypotenuse.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle. The radius is its hypotenuse.
- Find half the chord. The perpendicular from the centre to a chord bisects the chord.AM = 48 ÷ 2 = 24 cm (the perpendicular from the centre to a chord bisects the chord)
- Use Pythagoras in triangle OMA. The angle at M is 90°, and the radius OA is the hypotenuse.OA2 = 72 + 242 = 49 + 576 = 625
- Square root. The radius is positive.OA = √625 = 25 cm
Check. The distance from the centre to B is the same: √(72 + 242) = 25. ✓
Watch out. Add the squares to get the hypotenuse.
Method. Use the perpendicular from the centre to the chord, then take the distance from the radius.
- Draw the picture. The surface of the water is a chord. The lowest point of the water is at the bottom of the circle, directly below the centre.Chord = 16 cm, radius = 10 cm
- Find the distance from the centre to the chord. The perpendicular from the centre bisects the chord, which gives a right-angled triangle with half the chord and the radius.Half the chord = 8 cm
Distance2 = 102 − 82 = 100 − 64 = 36
Distance = 6 cm - Work out the depth. The depth is the radius take away the distance from the centre to the chord.Depth = 10 − 6 = 4 cm
Check. Water is 4 cm deep and the centre is 6 cm above the surface: 4 + 6 = 10 cm, the radius. ✓
Watch out. The depth is not the distance from the centre to the chord. It is the radius minus that distance.
Method. Use the perpendicular from the centre to the chord, then take the distance from the radius.
- Draw the picture. The surface of the water is a chord. The lowest point of the water is at the bottom of the circle, directly below the centre.Chord = 8 cm, radius = 5 cm
- Find the distance from the centre to the chord. The perpendicular from the centre bisects the chord, which gives a right-angled triangle with half the chord and the radius.Half the chord = 4 cm
Distance2 = 52 − 42 = 25 − 16 = 9
Distance = 3 cm - Work out the depth. The depth is the radius take away the distance from the centre to the chord.Depth = 5 − 3 = 2 cm
Check. Water is 2 cm deep and the centre is 3 cm above the surface: 2 + 3 = 5 cm, the radius. ✓
Watch out. The depth is not the distance from the centre to the chord. It is the radius minus that distance.
Method. Use the perpendicular from the centre to the chord, then take the distance from the radius.
- Draw the picture. The surface of the water is a chord. The lowest point of the water is at the bottom of the circle, directly below the centre.Chord = 16 cm, radius = 17 cm
- Find the distance from the centre to the chord. The perpendicular from the centre bisects the chord, which gives a right-angled triangle with half the chord and the radius.Half the chord = 8 cm
Distance2 = 172 − 82 = 289 − 64 = 225
Distance = 15 cm - Work out the depth. The depth is the radius take away the distance from the centre to the chord.Depth = 17 − 15 = 2 cm
Check. Water is 2 cm deep and the centre is 15 cm above the surface: 2 + 15 = 17 cm, the radius. ✓
Watch out. The depth is not the distance from the centre to the chord. It is the radius minus that distance.
Method. Find how far each chord is from the centre, then think about whether the chords are on the same side of the centre or on opposite sides.
- Chord of length 16 cm. The perpendicular from the centre bisects it, so half the chord is 8 cm. The radius is the hypotenuse.d12 = 102 − 82 = 100 − 64 = 36, so d1 = 6 cm
- Chord of length 12 cm. Half the chord is 6 cm.d22 = 102 − 62 = 100 − 36 = 64, so d2 = 8 cm
- Same side of the centre. If both chords are on the same side of the centre, subtract the distances.8 − 6 = 2 cm
- Opposite sides of the centre. If the chords are on opposite sides, add the distances.8 + 6 = 14 cm
Check. Both answers make sense: the circle has diameter 20 cm and 14 cm is less than that. ✓
Watch out. There are two possible pictures. Parallel chords can be on the same side of the centre or on opposite sides.
Method. Use the perpendicular from the centre and Pythagoras to find the distance from the centre to the chord.
- Find half the chord. The perpendicular from the centre to a chord bisects the chord.half the chord = 20 ÷ 2 = 10 cm
- Use Pythagoras. The radius is the hypotenuse.d2 = 102 − 102 = 0, so d = 0
- Conclude. A distance of 0 means the chord passes through the centre.So the chord is a diameter (its length is 2 × 10 = 20 cm). ✓
Check. The diameter of this circle is 2 × 10 = 20 cm, the same as the chord. ✓
Watch out. The diameter is the longest possible chord.
Method. Use the isosceles triangle made by two radii.
- Spot the isosceles triangle. OA and OB are radii so OA = OB.triangle OAB is isosceles with apex O
- Use the midpoint. The line from the apex to the midpoint of the base splits the triangle into two triangles with all three sides equal (OA = OB, AM = MB, OM common).the two triangles are congruent (SSS)
- Equal angles at M. They are matching angles of congruent triangles and they add up to 180° on the line AB.angle OMA = 180° ÷ 2 = 90°
Check. A numerical test: radius 13, AM = 12, OM = 5 satisfies Pythagoras: 25 + 144 = 169. ✓
Watch out. This is the converse of the main theorem.
Exercise 26I Chords, tangents and trigonometry (calculator allowed)
Calculator allowedMethod. Half the chord and the distance from the centre are the two short sides of a right-angled triangle whose hypotenuse is the radius.
- Find half the chord. The shortest distance from the centre meets the chord at 90° and bisects it.AM = 10 ÷ 2 = 5 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. OA is the radius and the hypotenuse.72 = OM2 + 52
- Rearrange. Subtract the known square.OM2 = 49 − 25 = 24
- Square root and round. Use the √ key, then round to 1 decimal place.OM = √24 = 4.899... = 4.9 cm (1 d.p.)
Check. 4.8992 + 52 = 49.00 and 72 = 49. ✓
Watch out. Keep all the decimal places on your calculator until the end, then round once.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle whose hypotenuse is the radius.
- Find half the chord. The shortest distance from the centre meets the chord at 90° and bisects it.AM = 13 ÷ 2 = 6.5 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. OA is the radius and the hypotenuse.92 = OM2 + 6.52
- Rearrange. Subtract the known square.OM2 = 81 − 42.25 = 38.75
- Square root and round. Use the √ key, then round to 1 decimal place.OM = √38.75 = 6.225... = 6.2 cm (1 d.p.)
Check. 6.2252 + 6.52 = 81.00 and 92 = 81. ✓
Watch out. Keep all the decimal places on your calculator until the end, then round once.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle whose hypotenuse is the radius.
- Find half the chord. The shortest distance from the centre meets the chord at 90° and bisects it.AM = 16 ÷ 2 = 8 cm (the perpendicular from the centre to a chord bisects the chord)
- Write Pythagoras for triangle OMA. OA is the radius and the hypotenuse.122 = OM2 + 82
- Rearrange. Subtract the known square.OM2 = 144 − 64 = 80
- Square root and round. Use the √ key, then round to 1 decimal place.OM = √80 = 8.944... = 8.9 cm (1 d.p.)
Check. 8.9442 + 82 = 144.00 and 122 = 144. ✓
Watch out. Keep all the decimal places on your calculator until the end, then round once.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle whose hypotenuse is the radius.
- Find half the chord. The perpendicular from the centre bisects the chord.AM = 9 ÷ 2 = 4.5 cm (the perpendicular from the centre to a chord bisects the chord)
- Use Pythagoras in triangle OMA. The radius OA is the hypotenuse, so add the squares.OA2 = 4.72 + 4.52 = 22.09 + 20.25 = 42.34
- Square root and round. Use the √ key, then round.OA = √42.34 = 6.507... = 6.5 cm (1 d.p.)
Check. The radius must be bigger than half the chord (4.5) and bigger than the distance (4.7). 6.5 is. ✓
Watch out. Keep all the decimal places on your calculator until the end, then round once.
Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle whose hypotenuse is the radius.
- Find half the chord. The perpendicular from the centre bisects the chord.AM = 18 ÷ 2 = 9 cm (the perpendicular from the centre to a chord bisects the chord)
- Use Pythagoras in triangle OMA. The radius OA is the hypotenuse, so add the squares.OA2 = 12.02 + 92 = 144.00 + 81 = 225.00
- Square root and round. Use the √ key, then round.OA = √225.00 = 15.000... = 15.0 cm (1 d.p.)
Check. The radius must be bigger than half the chord (9) and bigger than the distance (12.0). 15.0 is. ✓
Watch out. Keep all the decimal places on your calculator until the end, then round once.
Method. Split the isosceles triangle OAB into two right-angled triangles with the perpendicular from O, use sine, then double the angle.
- Draw the perpendicular from O to the chord. Let it meet AB at M. The perpendicular from the centre bisects the chord and also bisects angle AOB, because triangle OAB is isosceles.AM = 10 ÷ 2 = 5 cm
- Choose sine in the right-angled triangle OMA. We know the opposite side AM and the hypotenuse OA (the radius), and want the angle at O. Opposite and hypotenuse means sine.sin(angle AOM) = AM ÷ OA = 5 ÷ 8
- Use the inverse sine on a calculator. The inverse of sin finds the angle.angle AOM = sin−1(5 ÷ 8) = 38.68°
- Double it. Angle AOM is only half of angle AOB.angle AOB = 2 × 38.68° = 77.4° (1 d.p.)
Check. Check with cosine: the distance from O to the chord is √(82 − 52) = 6.24 cm, and cos−1(6.24 ÷ 8) = 38.68°. ✓
Watch out. Do not forget to double: the right-angled triangle gives only half of angle AOB. Keep extra decimal places until the end.
Method. Split the isosceles triangle OAB into two right-angled triangles with the perpendicular from O, use sine, then double the angle.
- Draw the perpendicular from O to the chord. Let it meet AB at M. The perpendicular from the centre bisects the chord and also bisects angle AOB, because triangle OAB is isosceles.AM = 9 ÷ 2 = 4.5 cm
- Choose sine in the right-angled triangle OMA. We know the opposite side AM and the hypotenuse OA (the radius), and want the angle at O. Opposite and hypotenuse means sine.sin(angle AOM) = AM ÷ OA = 4.5 ÷ 6
- Use the inverse sine on a calculator. The inverse of sin finds the angle.angle AOM = sin−1(4.5 ÷ 6) = 48.59°
- Double it. Angle AOM is only half of angle AOB.angle AOB = 2 × 48.59° = 97.2° (1 d.p.)
Check. Check with cosine: the distance from O to the chord is √(62 − 4.52) = 3.97 cm, and cos−1(3.97 ÷ 6) = 48.59°. ✓
Watch out. Do not forget to double: the right-angled triangle gives only half of angle AOB. Keep extra decimal places until the end.
Method. Split the isosceles triangle OAB into two right-angled triangles with the perpendicular from O, use sine, then double the angle.
- Draw the perpendicular from O to the chord. Let it meet AB at M. The perpendicular from the centre bisects the chord and also bisects angle AOB, because triangle OAB is isosceles.AM = 14 ÷ 2 = 7 cm
- Choose sine in the right-angled triangle OMA. We know the opposite side AM and the hypotenuse OA (the radius), and want the angle at O. Opposite and hypotenuse means sine.sin(angle AOM) = AM ÷ OA = 7 ÷ 12
- Use the inverse sine on a calculator. The inverse of sin finds the angle.angle AOM = sin−1(7 ÷ 12) = 35.69°
- Double it. Angle AOM is only half of angle AOB.angle AOB = 2 × 35.69° = 71.4° (1 d.p.)
Check. Check with cosine: the distance from O to the chord is √(122 − 72) = 9.75 cm, and cos−1(9.75 ÷ 12) = 35.69°. ✓
Watch out. Do not forget to double: the right-angled triangle gives only half of angle AOB. Keep extra decimal places until the end.
Method. The radius meets the tangent at a right angle, so triangle OAT is right-angled and we can use Pythagoras' theorem.
- Find the right angle. A tangent is perpendicular to the radius at the point of contact.Angle OAT = 90° (a tangent is perpendicular to the radius at the point of contact)
- Write Pythagoras for triangle OAT. The longest side is OT, opposite the right angle.OT2 = OA2 + TA2
92 = 42 + TA2 - Rearrange. Take the known square away from both sides.TA2 = 81 − 16 = 65
- Take the square root. This is not a whole number, so leave it as a surd or round it.TA = √65 = 8.1 cm (1 d.p.)
Check. 42 + 65 = 81 = 92. ✓
Watch out. OT is the hypotenuse, so it is the one you square and then subtract from, not add to.
Method. The radius meets the tangent at a right angle, so triangle OAT is right-angled and we can use Pythagoras' theorem.
- Find the right angle. A tangent is perpendicular to the radius at the point of contact.Angle OAT = 90° (a tangent is perpendicular to the radius at the point of contact)
- Write Pythagoras for triangle OAT. The longest side is OT, opposite the right angle.OT2 = OA2 + TA2
112 = 62 + TA2 - Rearrange. Take the known square away from both sides.TA2 = 121 − 36 = 85
- Take the square root. This is not a whole number, so leave it as a surd or round it.TA = √85 = 9.2 cm (1 d.p.)
Check. 62 + 85 = 121 = 112. ✓
Watch out. OT is the hypotenuse, so it is the one you square and then subtract from, not add to.
Method. The radius meets the tangent at 90°, so triangle OAT is right-angled. Use trigonometry to find the angle.
- Find the right angle. A tangent is perpendicular to the radius at the point of contact.angle OAT = 90° (the angle between a tangent and a radius is 90°)
- Choose the ratio. Relative to the angle T, the side OA is opposite and OT is the hypotenuse. Opposite and hypotenuse means sine.sin(angle ATO) = OA ÷ OT = 6 ÷ 10 = 0.6
- Use the inverse sine. The inverse of sin finds the angle on a calculator.angle ATO = sin−1(0.6) = 36.87...° = 36.9° (1 d.p.)
Check. Use Pythagoras to get AT = 8 cm, then tan(angle) = 6 ÷ 8 = 0.75 and tan−1(0.75) = 36.9°. ✓
Watch out. Check your calculator is in degree mode.
Method. Find the height of the isosceles triangle OAB using the perpendicular from O, then use area = ½ × base × height.
- Find half the chord. The perpendicular from the centre bisects the chord.half the chord = 13 ÷ 2 = 6.5 cm
- Find the height OM. Pythagoras in triangle OMA: the radius is the hypotenuse.OM2 = 8.52 − 6.52 = 72.25 − 42.25 = 30
OM = √30 = 5.477 cm - Work out the area. The base is the chord AB = 13 cm.area = ½ × 13 × 5.477 = 35.60 cm2 = 35.6 cm2 (3 s.f.)
Check. Another way: area = ½ × 8.5 × 8.5 × sin(AOB) where AOB = 2 × sin−1(6.5 ÷ 8.5) = 99.76°, which gives 35.60 cm2. ✓
Watch out. Use the exact value √30 in your calculator to avoid rounding errors.