Key ideas
- A line from the centre, O, that is perpendicular to a chord cuts the chord in half. It goes through the midpoint, M.
- So triangle OMA is right-angled at M, with
- OA = the radius (the hypotenuse)
- AM = half the chord
- OM = the distance from the centre to the chord
- Know two of these and use Pythagoras to find the third. Finding the radius? Add the squares. Finding a shorter side? Subtract.
- A tangent meets a radius at 90°, so a tangent, a radius and a line to the centre also make a right-angled triangle.
- Always draw the right-angled triangle on the diagram and mark the lengths you know, with the right angle. In the exam, the first mark is for doing exactly this: marshalling the facts in a right-angled triangle, with the right angle marked. The next mark is for a Pythagoras statement that matches your triangle.
- The marks usually go: •1 marshal the facts and recognise the right angle, •2 a Pythagoras statement that matches your triangle, •3 calculate the third side, •4 calculate what the question asks for (the chord, depth, width or height).
- A correct answer with no working gets no marks. If you don't draw the triangle, a correct Pythagoras statement can still earn •1 and •2, but a wrong one loses both. Sketch the triangle!
Symmetry in a circle
AB is a chord of the circle, centre O. OM is perpendicular to AB. AB = 18 cm. State the length of AM and the size of angle OMA.(2 marks)
Spot the symmetry.AM = 182 = 9 cm✓1
In plain EnglishA line from the centre that meets a chord at a right angle cuts the chord exactly in half.
The angle.Angle OMA = 90°✓2
In plain EnglishOM is perpendicular to AB, so it meets it at 90°. That is the right angle we need for Pythagoras.
Distance from the centre
A circle, centre O, has radius 10 cm. AB is a chord of length 16 cm. Calculate the distance from O to the chord (the length OM).(3 marks)
Marshal the facts.AM = 162 = 8 cmTriangle OMA: right angle at M, OA = 10 cm, AM = 8 cm✓1
On the diagram
In plain EnglishOM is perpendicular to AB, so it cuts the chord in half. Sketch the right-angled triangle OMA, mark the right angle and write on the lengths you know. That is the first mark.
Pythagoras.OM² = OA² − AM²OM² = 10² − 8²✓2OM² = 100 − 64 = 36
In plain EnglishOA is the radius, which is the hypotenuse. We want a shorter side, so subtract the squares.
Square root.OM = 36 = 6 cm✓3
On the diagram
In plain EnglishTake the square root to get the length.
⚠ Halve the chord first. The right-angled triangle uses half the chord, not all of it.
Length of a chord
A circle, centre O, has radius 8.5 cm. The chord AB is 4 cm from the centre. Calculate the length of the chord AB.(4 marks)
Marshal the facts, then Pythagoras.Triangle OMA: right angle at M, OA = 8.5 cm, OM = 4 cm✓1AM² = OA² − OM²AM² = 8.5² − 4²✓2AM² = 72.25 − 16 = 56.25
On the diagram
In plain EnglishTriangle OMA is right-angled at M. The radius OA is the hypotenuse, so subtract to find the shorter side AM.
Square root.AM = 56.25 = 7.5 cm✓3
In plain EnglishThis is only half the chord.
Double it.AB = 2 × 7.5 = 15 cm✓4
On the diagram
In plain EnglishM is the midpoint, so the whole chord is twice AM.
⚠ Don't stop at 7.5 cm. That is AM. The question asks for the whole chord AB.
Finding the radius
A chord AB of length 30 cm is 9 cm from the centre, O, of a circle. Calculate the radius of the circle. Give your answer to 1 decimal place.(3 marks)
Marshal the facts.AM = 302 = 15 cmTriangle OMA: right angle at M, AM = 15 cm, OM = 9 cm✓1
On the diagram
In plain EnglishOM cuts the chord in half, so AM = 15 cm. Put the facts you know into the right-angled triangle OMA.
Pythagoras.OA² = AM² + OM²OA² = 15² + 9²✓2OA² = 225 + 81 = 306
In plain EnglishThe radius OA is the hypotenuse, so this time we add the squares.
Square root and round.OA = 306 = 17.492…r = 17.5 cm✓3
In plain EnglishRound to 1 decimal place at the very end.
Pipes and tunnels
A pipe has a circular cross-section with radius 34 cm. The water surface, AB, is 56 cm wide. Calculate the depth of the water, d. Give your answer to 1 decimal place.(4 marks)
Marshal the facts.AM = 562 = 28 cmTriangle OMA: right angle at M, OA = 34 cm, AM = 28 cm✓1
On the diagram
In plain EnglishThe centre O is above the water. Draw OM down to the surface: it cuts AB in half.
Pythagoras for OM.OM² = OA² − AM²OM² = 34² − 28²✓2OM² = 1156 − 784 = 372OM = 372 = 19.287…✓3
In plain EnglishOA is the radius, the hypotenuse. Keep the full value on your calculator.
Depth.d = 34 − 19.287…d = 14.7 cm✓4
On the diagram
In plain EnglishFrom the centre down to the bottom of the pipe is a radius, 34 cm. The water is that radius minus OM.
Check the picture. If the centre is above the water, depth = radius − OM. If the water covers the centre, depth = radius + OM.
A tunnel has a circular cross-section with radius 5 m. The flat floor, AB, is 8 m wide. Calculate the height, h, of the tunnel.(4 marks)
Marshal the facts.AM = 82 = 4 mTriangle OMA: right angle at M, OA = 5 m, AM = 4 m✓1
On the diagram
In plain EnglishOM goes from the centre down to the floor at a right angle, so it cuts the floor in half.
Pythagoras for OM.OM² = OA² − AM²OM² = 5² − 4²✓2OM² = 9, so OM = 9 = 3 m✓3
In plain EnglishOA is a radius of 5 m.
Height.h = 5 + 3 = 8 m✓4
On the diagram
In plain EnglishThe tunnel goes up from the floor to O (3 m) and then a full radius from O to the top (5 m). This time we add.
Tangents
PT is a tangent to the circle, centre O, at T. The radius is 5 cm and PT = 12 cm. Calculate the length of OP.(3 marks)
Marshal the facts.Angle OTP = 90° (tangent and radius)Triangle OTP: right angle at T, OT = 5 cm, TP = 12 cm✓1
On the diagram
In plain EnglishA tangent always meets the radius at 90°, so triangle OTP is right-angled at T and OP is the hypotenuse.
Pythagoras.OP² = OT² + TP²OP² = 5² + 12²✓2OP² = 25 + 144 = 169
In plain EnglishOP is the hypotenuse, so add the squares.
Square root.OP = 169 = 13 cm✓3
In plain EnglishTake the square root.
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Past paper questions
Ready for exam-style questions? Pythagoras in a circle comes up in National 5 almost every year.
National 5 Pythagoras: Pythagoras in a circleThe maths.scot Pythagoras page lists every National 5 "Pythagoras in a circle" question (2014 onwards) with worked videos.maths.scot Open ↗ (opens in a new tab) Essential Skills 27: circle calculations (Pythagoras)A practice sheet of chord, depth and radius questions.maths.scot · PDF Open ↗ (opens in a new tab) Essential Skills answersAnswers for Essential Skills 25 to 28.maths.scot · PDF Open ↗ (opens in a new tab) Standard Grade Credit: Pythagoras in a circleOlder exam questions, very similar in style.maths.scot · PDF Open ↗ (opens in a new tab) Standard Grade Credit answersAnswers for the questions above.maths.scot · PDF Open ↗ (opens in a new tab)