Unit 2
6 marks · HardIn the diagram below, points A and B lie on the circumference of the circle with centre O. The straight line AB is a chord, and OC is a radius perpendicular to AB, meeting AB at M. The length of OC is 7 cm and the distance from O to M is 4 cm.
Calculate the length MC.
Find the length AB.
Express your answer to part (b) in standard form.
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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(a) √(7^2 - 4^2) = √(49 - 16) = √33 ≈ 5.74 cm · must show correct Pythagoras' calculation
(b) 2 × 5.74 = 11.48 cm · must show doubles MC to get length of chord AB
(c) 1.15 × 10^1 cm · must show Expresses to 3 sig. fig. in standard form
Must show: Correct use of right triangle OMC, Doubling MC for AB
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Sets up correct right triangle and applies Pythagoras’ theorem for MC.
A1 (a) · 1 mark · Obtains correct value for MC within allowed tolerance (5.74 cm). (follow-through)
M2 (b) · 1 mark · Recognises AB = 2 × MC and carries out calculation. (follow-through)
A2 (b) · 2 marks · Correct answer for AB (accept within 0.1), with units. (follow-through)
A3 (c) · 1 mark · Converts answer for AB to standard form, 3 sig. fig. (follow-through)
Worked solution
a) In right triangle OMC, OC = 7 cm, OM = 4 cm, so: MC = √(OC² - OM²) = √(49 - 16) = √33 ≈ 5.7446 cm ≈ 5.74 cm to 2 d.p.
b) Since OC is perpendicular bisector of chord AB, AB = 2 × MC = 2 × 5.74 = 11.48 cm
c) In standard form: 11.48 = 1.148 × 10^1 cm = 1.15 × 10^1 cm (to 3 sig. fig.)