A circular cake tin of radius 5 cm sits on a bench. A ribbon is pulled taut from an anchor point P on the bench so that it just touches the tin at point T. ...
Question 1 Report
A circular cake tin of radius 5 cm sits on a bench. A ribbon is pulled taut from an anchor point P on the bench so that it just touches the tin at point T. The distance OP from the centre O of the tin to the anchor point is 13 cm.
- State, giving a reason, the size of angle OTP. (1)
- Work out the length of the ribbon, PT. (2)
Answer Details
A tangent line touches a circle at exactly one point and is always perpendicular to the radius drawn to that point, which turns the ribbon, the radius and the line OP into a right-angled triangle solvable by Pythagoras.
(a) Since the ribbon PT is tangent to the tin at T, angle OTP \(=90^\circ\), because a tangent is always perpendicular to the radius at the point of contact [1 mark].
(b) With the right angle at T, OP is the hypotenuse of triangle OTP, so Pythagoras gives \( PT=\sqrt{OP^2-OT^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12 \) cm [2 marks].
The tangent-radius right angle is what makes this a straightforward Pythagoras problem rather than needing trigonometry; recognising a tangent whenever a line "just touches" a circle is the key first step.
Download The App On Google Playstore
Everything you need to excel in your exams
