A cylindrical water tank on a farm lies on its side, with circular cross-section and centre \(O\). Rainwater inside forms a chord \(AB\) across the circle, ...
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A cylindrical water tank on a farm lies on its side, with circular cross-section and centre \(O\). Rainwater inside forms a chord \(AB\) across the circle, and \(M\) is the midpoint of \(AB\). A dipstick shows \(AB = 24\) cm and \(OM = 5\) cm.
- Write down the size of angle \(OMA\). (1)
- Work out the radius of the tank. (3)
Answer Details
A line from the centre of a circle to the midpoint of a chord always meets it at right angles, which is exactly what makes it possible to find the radius from a chord length and the distance to its midpoint.
(a) Angle \(OMA = 90^\circ\), because the perpendicular from the centre of a circle to a chord always bisects the chord and meets it at right angles. [1 mark]
(b) Since \(M\) is the midpoint of chord \(AB\), \(AM = 24 \div 2 = 12\) cm. Triangle \(OAM\) is right-angled at \(M\), with the radius \(OA\) as the hypotenuse, so Pythagoras' theorem gives \(OA^2 = OM^2 + AM^2 = 5^2 + 12^2 = 25 + 144 = 169\), so the radius \(OA = \sqrt{169} = 13\) cm. [3 marks]
Exam tip: this is a \(5\text{-}12\text{-}13\) Pythagorean triple in disguise; recognising the pattern once \(OM = 5\) and \(AM = 12\) are known lets you write down \(OA = 13\) directly, as a check on the square-root calculation.
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