A nice puzzle
Hi...!!!
A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
posted by Swati Gupta at 8:27 PM
17 Comments:
That's a really interesting question..still thinking...
THIS PUZZLES ME TOO
I HAVE ANOTHER QUESTION IF THE MINUTE HAND OVERTAKES THE HOUR HAND EVERY 65 MIN IS IT GAINING OR LOSING TIME AND BY HOW MUCH
i too want to post questions tell me what to do?
First reveal your identity(newton)
you wont know me as i am in class x
anyway i am abhishek nanda x-g.is that fine?
Abhishek, your identity is fine with us but we do need your e-mail address inorder to invite you to be a member of this blog, on accepting which you too will be able to post questions.
As for your question, the clock loses time by 5 minutes everytime the minute hand overtakes the hour hand.
well the answer is not right?try it again my id
abhishek_force@yahoo.com
abhishek.nanda@gmail.com
I have sent you an invitation, and I am quite sure that the clock loses time, although the number might not be correct.
no the clock is gaining time.here is the the explanation:
the minute hand covers 6degrees every minute and the hour hand 1/2 degree every minute
say between 1 and 2 o'clock the hands would be together after x min
we have,
6x=1/2x + 30
since the hour hand is already at 1 or 30 degrees ahead.
x=60/11
time is 1 + 60/11
say between 2 and 3 o'clock the hands would be together after y min
6x=1/2x + 60
x=120/11
time = 2+120/11
subtracting,
1hr +60/11min
60 min +5min +5/11min
65 + 5/11 min
therefore the the minute hand is overtaking the hour one in 65min instead of 65+5/11 min and is hence gaining 5/11 min
Good you accepted the invitation. About your solution, its fine.
The clock mentioned in your question has got either a faulty minute hand or a faulty hour hand. What you have assumed is that the minute hand of your clock moves faster, which means the clock is gaining time but what I had assumed was the hour hand to be slower, and if the time shown by the clock would be measured with respect to the position of the hour hand, the clock appears to be losing time.
Ok..what about the original question???
Right, I'll get to that once I'm done with this horrendus upcoming Chemistry board exam.
all the best!!!
Well Swati, I thought a lot but couldn't figure out. Can we have the answer please?
sorry for the delay...
the answer is sqrt((1+sqrt5)/2)..
It is a little difficult to explain the solution here..but let's try..
a'' : (a) square
a^ : root of (a)
/\ : triangle
let the rectangle be ABCD with AD=a and AB=b, such that a<=b
let the fold be EF where E is on AB and F on CD such that EF is of length x..and the diagonal is of length d..let EF and BD intersect at 0.
By symmetry, EF and BD bisect each other at right angles.(coz once you fold the rectangle, both the sides would look exactly similar)..
SO, OB=d/2 and OE=x/2.
Now, /\OEB is similar to /\DAB
so, by equal ratios: a/b=x/d
but d=(a''+b'')^
so, x=a/b*(a''+b'')^
and A/Q x=b;
which means :
b''=a''(a''+b'')
solve the biquadratic for b and get the ratio of the sides...
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