Unit 6
4 marks · Normal4. (a) The diagram below shows a circle with centre O and points A, B, C, and D on the circle. AC and BD intersect at E inside the circle.
If AE = 2.5 cm, EC = 3.5 cm, BE = 4 cm, calculate the length of ED.
Write, using set notation, the set S of all possible real values ED can take if all other lengths are positive and fixed.
Suppose the lengths above are in direct proportion to the radius r of the circle, currently 4 cm. Write an expression for ED when the radius of the circle is doubled.
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.
Show answer
(a) 2.188 cm
(b) S = { x | x > 0 }
(c) ED = 2 × 2.188 = 4.376 cm (or 2×ED) cm
Must show: Correct use of intersecting chord theorem
Common mistakes: Wrong application of theorem; Forgetting to scale by radius
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Use intersecting chord theorem correctly: AE × EC = BE × ED.
A1 (a) · 1 mark · Obtain correct value for ED (2.19 cm). (follow-through)
B1 (b) · 1 mark · States correct set notation for possible ED (>0).
A2 (c) · 1 mark · Scales ED directly with r, doubles it if r is doubled.
Worked solution
a) By intersecting chord theorem: AE × EC = BE × ED. So, 2.5 × 3.5 = 4 × ED. ED = (2.5 × 3.5) / 4 = 8.75 / 4 = 2.1875 cm.
b) Since ED > 0 (lengths must be positive), S = {x | x > 0}.
c) If r doubles, then by direct proportion, ED new = 2 × 2.188 ≈ 4.376 cm.