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BdMO(Divisional)

Posted: Sat Dec 11, 2010 12:22 pm
by rakeen
Can you hlp me with these?

Re: BdMO(Divisional)

Posted: Sat Dec 11, 2010 12:28 pm
by rakeen
Another one:

Re: BdMO(Divisional)

Posted: Sat Dec 11, 2010 12:35 pm
by Moon
Please create separate posts for each problem from the next time, and use better topic. No one might be interested in your topic if they just see "BdMO divisional" in your topic. I mean posting in this forum already made it it clear this the problems are from BdMO divisional.

Re: BdMO(Divisional)

Posted: Sat Dec 11, 2010 4:01 pm
by rakeen
thnx for the RULES. But where's the solution! :evil:

Re: BdMO(Divisional)

Posted: Sat Dec 11, 2010 11:05 pm
by Moon
rakeen wrote:thnx for the RULES. But where's the solution! :evil:
Whoa! Slowly bro! Please be a bit patient. Someone will solve your problem. Anyway it is better to write the statement of your problem in the forum and attach the image.
It is hard to read most of the statements. (You hand writing is better than that of me, but the image quality are not good)

Re: BdMO(Divisional)

Posted: Sun Dec 12, 2010 12:51 am
by TIUrmi
Hints:

To problem 1:
(Ratio of heights or sides)^3 = Ratio of volumes.

Problem 2:
Probably you have made some typos or linguistic mistakes in stating the problem. If you find it difficult to translate a problem into English you may write it in Bengali.
The problem, I think, is wrong because a cylinder inside a sphere can never have radius equal to that of the sphere. You may wonder why? Look, the upper and lower base of cylinder is parallel, so a line that goes just midway along the cylinder will be a diameter of the sphere and the midpoint of it is the center of the sphere, say $O$. Join $O$ to the circumference (পরিধি) of the base of the cylinder at $P$. Drop a perpendicular from $O$ to the the base at $R$; I hope you understand $OP$ is the hypotenuse (অতিভূজ) of a right angle triangle (সমকোণী ত্রিভূজ) whose base (ভূমি) is $OR$, the radius of the base of the cylinder. Now $OP$ can never be equal to $OR$.

Problem 3:

$\triangle PAB$ and $\triangle PCD$ are similar and $\frac {\triangle PAB}{\triangle PCD} = \frac {YA}{XR}$

Problem 4:

Note that the cone has two similar triangle inside it. One is the cross section of the whole cone and another is the cross section of the part of the cone inside the cylinder. Find heights and volume of the different sections.

Try solving them now. Have I mentioned something that you don't know already?

Re: BdMO(Divisional)

Posted: Sun Dec 12, 2010 3:59 pm
by rakeen
It was made by Microsoft Paint. And I made the resolution low, coz it would have been better for me to upload it. But now I can c it was too small! And the font was by bijoy and unicode.
Why do you work so hard when you can just write it on the forum?? :)
Anyway when you add texts to images the font quality gets worse with the quality of images reduced.

Re: BdMO(Divisional)

Posted: Sun Dec 12, 2010 4:07 pm
by rakeen
TIUrmi wrote in Problem 3: XR




where did u get the R?

Re: BdMO(Divisional)

Posted: Sun Dec 12, 2010 6:56 pm
by rakeen
@TIUrmi:

oops! the raius is half of the sphere. So the problem can be solved by Pythagoras. thnx.

Re: BdMO(Divisional)

Posted: Sun Dec 12, 2010 7:07 pm
by rakeen
Prob 3.

So what. I don't know the other 'bahu' of each triangle.