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## Question

secondary 4 | A Maths

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##### Sonia

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Date Posted: 1 week ago
Views: 7

### See 5 Answers

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##### Eric Nicholas K's answer
2445 answers (Tutor Details)
Hi Sonia! Here is my version of the answers for part (i).
Date Posted: 1 week ago
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##### Eric Nicholas K's answer
2445 answers (Tutor Details)
Part (iii), halfway done
Date Posted: 1 week ago
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##### Eric Nicholas K's answer
2445 answers (Tutor Details)
Hi Sonia! I’m sorry for my delayed post. Here is the continuation of my part (iii) working, using trigonometry which the other two tutors have not used. You can use this approach if you wish.
Date Posted: 1 week ago
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1868 answers (Tutor Details)
1st
Last part not difficult, just v tedious
Date Posted: 1 week ago
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57 answers (Tutor Details)
I have added the first portion on how to solve it without assuming the mid point. Just an alternative which I thought is interesting.
Date Posted: 1 week ago
Jiayang
1 week ago
It's not an assumption it's a property of circles. The perpendicular bisector of any chord passes through the centre. I used to do it your way when I was in school haha
Bendy
1 week ago
Oh no, don't get me wrong. Because it wasn't stated that the points are opposite, so I thought it wasn't as suitable to just directly get the mid point using that formula, as it might be 2 points forming a chord. But as you mention, it is a coincidence.

I do agree with you that the perpendicular bisector is the more efficient method. I tried doing with the radius method for part iii and it was hell. Great way of solving.
Eric Nicholas K
1 week ago
The proof of this property requires some congruency of triangles and some reasoning.
Eric Nicholas K
1 week ago
There is yet a third method which involves several simultaneous equations, so if a student does not realise this property, he/she can still solve the question based on other approaches.