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secondary 4 | E Maths
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Hi I need help with part a of this qns only, thks:) hope to receive a response asap
Date Posted: 5 years ago
Views: 239
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Part a iii is probably a tough one. This requires you to make length comparisons based on similar triangles.
Date Posted:
5 years ago
Icic, thks for the explanation. Sry but do u mind helping me with part b(v) and b(vi) too cas I seem to be having troubles with these too, thkiew:)
The idea for part b is this.
There are two kinds of pairs of triangles which you will see in part b.
Case 1: Triangles have same height, but different base (or the other way round). Their area ratio is the same as their base ratio (or height ratio if height is the one different, but most of the time we take the different one to be the base).
Case 2. The two triangles which we have picked out are similar triangles. Their area ratio is the square of their length ratio.
There are two kinds of pairs of triangles which you will see in part b.
Case 1: Triangles have same height, but different base (or the other way round). Their area ratio is the same as their base ratio (or height ratio if height is the one different, but most of the time we take the different one to be the base).
Case 2. The two triangles which we have picked out are similar triangles. Their area ratio is the square of their length ratio.
If I have time later (and I can still remember), I will do up all the sub questions in part b.
Ok, thks a lot
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You have to correctly identify whether the two triangles which you are trying to compare are similar, or they have same bases but different heights, or they have different bases but same heights.
In some cases, as is the case for parts v and vi, I have to split he quadrilateral into two smaller triangles for easier computation and visualisation.
In some cases, as is the case for parts v and vi, I have to split he quadrilateral into two smaller triangles for easier computation and visualisation.
Date Posted:
5 years ago
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