tag:blogger.com,1999:blog-2787161678592839240.post6188941154965898510..comments2024-03-29T12:30:29.739+05:30Comments on www.mathsblog.in : Maths Blog for School Teachers & Students: ചന്ദ്രക്കലയുടെയും സമചതുരത്തിന്റെയും വിസ്തീര്ണ്ണം തുല്യമാണ്...വി.കെ. നിസാര്http://www.blogger.com/profile/14303804236214732024noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-2787161678592839240.post-80254284525922026242009-08-25T21:13:24.989+05:302009-08-25T21:13:24.989+05:30THE answer can be proved using Pythegorean preposi...THE answer can be proved using Pythegorean preposition.step 1:THE semi circle formed on the hypotonues is equalto the sum of the area of the semi circles formed on the two other sides<br />step 2: throwing the big semi circle over to the other side we see two crescents taken together are equal to the triangle .<br />step 3:if we take an isoceles triangle then each of these two crescents will be equal to the triangle.<br />step:4: hence it is possible to form a right isoceles triangle whose area will be equal to that of a crescent and we can form a square with this triangle. SO THE AREA OF BOTH SQUARE AND CRESCENT ARE EQUAL( picture is also sent for clarification.if the blog team publish it is also easy to understand...vijayan N M (KPMSMHS ARIKKULAM)vijayanhttps://www.blogger.com/profile/02345885159669791726noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-36598787929814808652009-08-25T20:08:14.845+05:302009-08-25T20:08:14.845+05:30Dear Muneer,
There is a logical argument behind th...Dear Muneer,<br />There is a logical argument behind this construction.<br />Step 1<br /> let us extend the diagonal CA first.The circle with center A and radius root 2 x meet at a point on C A extented.Now we can imagine a semicircle clearly.<br />Step 2<br /> Think of the circle with radius x and center B. It will meet at a point on the semicircle that I draw previously.It is logically true.Let the meeting point is E.<br />step 3<br /> I think you ask the clarification (logical) about the colliniarity of C B and E<br />ABE IS A RT TRIANGLE.ABC is also a rt triangle.They are congruent and together form isocilus rt triangle EAC.we can complete a square by imaging another triangle congruent to it and joining hypotenuseof AEC.<br />This rt triangle will have a circumcircle.The part of that circumcircle is not just an arc,it is semicircle itself as EC is circumdiametre or hypotenuse<br /><br />JOHN P A HIBHS VARAPUZHAAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-611299779922085452009-08-25T13:09:06.871+05:302009-08-25T13:09:06.871+05:30Muneer, u r right...
But u pls read the Answer ca...Muneer, u r right... <br />But u pls read the Answer carefully.<br />they are indicated it too...Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-65827745094725987612009-08-25T12:09:19.684+05:302009-08-25T12:09:19.684+05:30ഒരു ചെറിയ സംശയം.
>> ഇപ്പോള് C,B,E എന്നീ ബിന...ഒരു ചെറിയ സംശയം.<br />>> ഇപ്പോള് C,B,E എന്നീ ബിന്ദുക്കള് ഒരു നേര്രേഖയിലാണ്<br /><br />ഈ നിഗമനം എന്ത് അടിസ്ഥാനത്തിലാണെന്ന് മനസ്സിലായില്ല. ചോദ്യത്തില് പറഞ്ഞിരിക്കുന്നത് "ബി കേന്ദ്രം ആയ ചാപം എന്നാണ്". "ബി കേന്ദ്രം ആയ അര്ദ്ധവൃത്തം" എന്ന് പറഞ്ഞിരുന്നെങ്കില് മുകളിലത്തെ നിഗമനം ശരി ആവുമായിരുന്നു.Muneerhttps://www.blogger.com/profile/12407187793935604761noreply@blogger.com