Ask Singapore Homework?

Upload a photo of a Singapore homework and someone will email you the solution for free.



Question

secondary 4 | A Maths
6 Answers Below

Anyone can contribute an answer, even non-tutors.

Answer This Question
Sonia
Sonia

secondary 4 chevron_right A Maths chevron_right Singapore

Please help me with this qns! Thanks a lot:)

Date Posted: 4 years ago
Views: 311

See 6 Answers

done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Hi Sonia! Here is my version of the answers for part (i).
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Part (iii), halfway done
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Eric Nicholas K
Eric Nicholas K's answer
5997 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.
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Jiayang
Jiayang's answer
1883 answers (Tutor Details)
1st
Last part not difficult, just v tedious
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Bendy
Bendy's answer
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.
Jiayang
Jiayang
4 years 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
Bendy
4 years 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
Eric Nicholas K
4 years ago
The proof of this property requires some congruency of triangles and some reasoning.
Eric Nicholas K
Eric Nicholas K
4 years 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.
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
sstrike
Sstrike's answer
5318 answers (A Helpful Person)