BAIHUI XU's answer to lana <3's Secondary 3 A Maths Singapore question.
hii! its helpful ty! but my school teacher did the ratio as PB=y/2 and BQ=x/2, i dont really understand how to derive this ? if the ratio of PB=y/2 and BQ=x/2 then what is the ratio for SD? And DR?
Hi, your teacher let AB = y and AD = x,
so PB = y/2, BQ = x/2,
in this case, SD = 3y/4, DR = 3x/4,
I do not like his method because it involved a lot of fractions and seemed not easy to understand.
But his idea is the same as mine.
so PB = y/2, BQ = x/2,
in this case, SD = 3y/4, DR = 3x/4,
I do not like his method because it involved a lot of fractions and seemed not easy to understand.
But his idea is the same as mine.
ah I see! thank you so much for clarifying! I appreciate it<3
hi again! just wondering, how to you derive the 3y/4 and the 3x/4? I’m not quite sure how you get that
Actually, Baihui made a slight oversight.
It's AR : AD (rather than AR : RD) which is 1 : 3.
A similar argument ensues for CS : CD.
It's AR : AD (rather than AR : RD) which is 1 : 3.
A similar argument ensues for CS : CD.
Lana, what Baihui meant was that based on the written pencil working (of letting AD = y and CD = x), then RD would be 2y/3 and SD would be 2x/3.
[RD takes up 2/3 of the length of AD while SD takes up 2/3 of the length of CD]
[RD takes up 2/3 of the length of AD while SD takes up 2/3 of the length of CD]
alright! I understand now! thank you very much! I really appreciate it<3
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