Shape and Space 6

4 marks · Hard

Circle with centre O, points Q, R, S and T.

In the diagram, O is the centre of the circle. QR and ST are two chords which intersect at right angles at point P inside the circle. If OP = 3 cm, QR = 6 cm, and ST = 8 cm, find the radius of the circle.

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

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5 cm (±0.05)

Also accept: 5; 5.0; 5 cm; 5.00; r = 5

Common mistakes: Forgetting OP is not radius; Using incorrect chord lengths; Not applying chord intersection property

Show mark scheme and worked solution

Marking points

M1 · 1 mark · Uses diameter or relation for intersecting chords property (QR × ST = ...).

M2 · 1 mark · Applies Pythagoras with OP, radius, and distance to chord. (follow-through)

A1 · 1 mark · Forms correct equation linking lengths. (follow-through)

A2 · 1 mark · Finds radius as 5 cm. (follow-through)

Worked solution

Let the radius be r. Since QR and ST are chords that intersect perpendicularly at P, and OP = 3 cm.

Let’s let P be at a distance 3 cm from the centre. Chord QR is 6 cm, so its midpoint is 3 cm from the centre (since intersecting chords at right angles through the centre split each other in half).

For chord QR: half-length = 3 cm, perpendicular distance from centre = 3 cm.

But need to check via the fact that both chords intersect at right angles, lengths split from P: QR: QP = RP = 3 cm, ST: SP = TP = 4 cm.