is there a typo error in last 2 lines
Yes, a copy-paste error. Corrected, thanks. Note to self, do not write maths posts when suffering from a cold 🙂 All sorts of minor errors crop up!
I want more of these kind of questions where we have to fit in a figure inside another and maximize it.There are plenty of scenarios. Please _/\_
As of now I have discussed mainly triangles so am a bit restricted. After finishing this series on triangles at some future point I will do some posts on circles and polygons, at that time I will be able to do more justice to these types of figures! I don’t know if it will be in time for CAT ’13 though 🙂
Dear J please increase the frequency and the content. Please I beg you. It can be a life saver for some of us. Please Please Please
Life saver for you pal….but I would probably die 🙂 As it is I am committing about 10-15 hours a week to this (over and above my normal duties at work…currently at peak season). If I try to do any more quality will drop drastically….
LOL , i understand. We can do one thing , if it goes with your flexibility , that you can post only Quant posts. eng we can do on our own 🙂 , but i certainly know that eng posts are a day saver for you 😉 🙂 . Over to you Sire.
Sir shouldnt it be OC is a median to PQ rather than Altitude?
It is an isosceles triangle and hence that median and altitude both will be the same…think about it!
May i know how OC =x/2
Isosceles right-angled triangle. Draw the median and convince yourself they will all be the same!
One way would be to take tan (45) =1 = opp/adjacent. OC would come out as x/2.
Correct me if I am wrong sir.
Yes, that’s right.
Solved this by taking x+y=1 and 2y^2=x^2 +1. Answer comes out 8-4rt3~1.07 and acc to this ~1.03. Where did this discrepancy arise?
P.S-Great blog supplementing my revision for C’14!
The error is of the same order of magnitude as the errors in approximating rt2 and rt3; so perfectly acceptable under the circumstances.
the equilateral triangle inside the square, from where did we get that from? or is it our own creation assuming it to be an equilateral triangle?
There will be an equilateral triangle possible in that orientation i.e. with a vertex common to that of the square (we need to visualise to realise this). Whether it will be larger than the other, well, that we have to check which is why we took both extreme cases.
*i am talking about the second case.
ok! got it! thank you! 🙂
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