tag:blogger.com,1999:blog-2787161678592839240.post2737383081967451079..comments2024-03-28T15:06:26.752+05:30Comments on www.mathsblog.in : Maths Blog for School Teachers & Students: ലംബകത്തിന്റെ പരപ്പളവ് (Area)വി.കെ. നിസാര്http://www.blogger.com/profile/14303804236214732024noreply@blogger.comBlogger15125tag:blogger.com,1999:blog-2787161678592839240.post-17378130635908405572010-01-16T09:42:21.397+05:302010-01-16T09:42:21.397+05:30let O be the poit of intersection of DB &AC
L...let O be the poit of intersection of DB &AC<br /><br />Let h1 be height to AB,&h2 be the height to DC from O<br /><br />since triangle OAB & OCD are similar<br />we have h1/h2=AB/DC=root a/rootb<br /><br />(AB+DC)/DC=(root a+root b)/root b (1)<br />(h1+h2)/h2=(root a+root b)/root b (2)<br />multiplying (1) & (2)<br />(h1+h2)(AB+DC)/h2 * DC =(root a+root b)^2/b<br /><br />Let h=h1+h2 distance between parallel lines <br />And by multipliying both numerator & denminator of LHS with 1/2<br />half h*(AB+DC)/half DC*h2 =(root a+root b)^/b<br /><br />therefor Area of trapezium =(root a +root b)^2<br />(half h2* DC=b)<br /><br />muraleedharan sir നോട് കടപ്പാട്<br />തോമസ് വി ടിsajan paulhttps://www.blogger.com/profile/11067627975897595499noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-3734463000178788952010-01-13T08:27:28.110+05:302010-01-13T08:27:28.110+05:30ഫോൺ നമ്പർ 240-2192 ആണു്. പക്ഷേ, ഇതു കണക്കുകൂട്ടലു...ഫോൺ നമ്പർ 240-2192 ആണു്. പക്ഷേ, ഇതു കണക്കുകൂട്ടലുകൾ വഴി കണ്ടുപിടിക്കാൻ എനിക്കു കഴിഞ്ഞില്ല. ഒരു കമ്പ്യൂട്ടർ പ്രോഗ്രാം ഉപയോഗിക്കേണ്ടി വന്നു.Umesh::ഉമേഷ്https://www.blogger.com/profile/06474085260183409502noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-74413211447203903302010-01-13T00:34:20.511+05:302010-01-13T00:34:20.511+05:30This comment has been removed by the author.Umesh::ഉമേഷ്https://www.blogger.com/profile/06474085260183409502noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-1506332584334120642010-01-13T00:23:21.886+05:302010-01-13T00:23:21.886+05:30To draw a circle with area equal to the sum of are...To draw a circle with area equal to the sum of areas of two given circles:<br /><br />Draw two concentric circles with the same area as the two given circles. (Use the construction given in <a href="http://www.usvishakh.net/documents/problems.pdf" rel="nofollow">this document</a>, Section 5.6) Draw a pair of perpendicular lines passing through the center. Let one line meet one circle at P and the other line meet the other circle at Q. With P as center and PQ as radius, draw a circle. The area of this circle will be the sum of the areas of the other two circles.Umesh::ഉമേഷ്https://www.blogger.com/profile/06474085260183409502noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-71935310908712873042010-01-13T00:13:42.883+05:302010-01-13T00:13:42.883+05:30I think the question John PA asked at January 12, ...I think the question John PA asked at January 12, 2010 7:43 PM is to construct, using straight edge and collapsing compass alone, a circle whose area is the some of the areas of two given circles. All the answers so far were analytical solutions.Umesh::ഉമേഷ്https://www.blogger.com/profile/06474085260183409502noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-38905912130441512692010-01-12T22:50:03.617+05:302010-01-12T22:50:03.617+05:30take any rt angled triangle. the area of the circl...take any rt angled triangle. the area of the circle with hypotanues as diameter .it is the sum of the area of circle drawn with other two sides as diameterVIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-20358060328481923452010-01-12T22:38:33.322+05:302010-01-12T22:38:33.322+05:30thanks umesh sir,muraii sir and bhama madam for on...thanks umesh sir,muraii sir and bhama madam for ontime response.<br /><br />MY TELEPHONE NUMBER IS CHANGED .BUT I forget To note it down .will u pl help me?<br />"it had a three digit prefix and the rest was another 4 digit number,like xxx-xxxx.I did remember that the prefix ,subtracted from half the square of the rest of the number gave me whole phone number as a result. "what is my new number?vijayanhttps://www.blogger.com/profile/02345885159669791726noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-680976046561560032010-01-12T22:18:51.450+05:302010-01-12T22:18:51.450+05:30Let Radius of small circles A & B
Radius of b...Let Radius of small circles A & B<br />Radius of big circle C<br /><br />Area pi A^2 + pi B^2 = Pi C^2<br /><br />Then A^2 + B^2 = C^2<br /><br />Radius of the big circle C = root(A^2 + B^2)bhamahttps://www.blogger.com/profile/04270776567671430504noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-29175624273448155742010-01-12T21:38:39.462+05:302010-01-12T21:38:39.462+05:30നിര്ദ്ദേശാങ്കജ്യമിതിയില് നിന്നും
REVISION MODULE...നിര്ദ്ദേശാങ്കജ്യമിതിയില് നിന്നും<br />REVISION MODULE 1<br /><a href="http://sites.google.com/site/schoolmathsgroup/revision1/moduleone?attredirects=0&d=1" rel="nofollow"> click here </a>JOHN P Ahttps://www.blogger.com/profile/02064365401403870252noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-4579901464768333002010-01-12T19:43:07.486+05:302010-01-12T19:43:07.486+05:30How can we construct a circle whose area is equal ...<a rel="nofollow"><br />How can we construct a circle whose area is equal to the sumof the areas of other two circles?</a>JOHN P Ahttps://www.blogger.com/profile/02064365401403870252noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-13663724641473253282010-01-12T19:20:25.418+05:302010-01-12T19:20:25.418+05:30Find the smallest right angled triangle
whose si...Find the smallest right angled triangle <br /><br />whose sides are integers <br /><br />The circumference is the square of an integer <br /><br />The area is the cube of an integer?AZEEZhttps://www.blogger.com/profile/07062507383328017955noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-20120870476396199092010-01-12T15:23:38.387+05:302010-01-12T15:23:38.387+05:30Let O be the meeting point of AC & DB
since OA...Let O be the meeting point of AC & DB<br />since OAB & OCD are similar AB/CD= root(A/B)<br />Draw a line through O parallel to AB & DC<br />Let h1 be the altitude from O to AB <br />& h2 be the altitude from O to DC<br />Now h1/h2 = AO/OC =AB/CD =root(A/B)<br />Therefore we can take h1 = (rootA)*x<br />& h2 = (rootB)*x<br />lllrly AB = (rootA)*y & DC =(rootB)*y<br />Since area of OAB = A <br />we get half*h1*AB = A<br />ie half*(rootA)*x*(rootA)*y = A<br />ie half*xy = 1<br />Required area =half*(h1+h2)*(AB+DC)<br /> =half*xy*((rootA)+(rootB))^2<br /> =((rootA)+(rootB))^2<br /><br /><br />MURALEEDHARAN.C.RMURALEEDHARAN.C.Rhttps://www.blogger.com/profile/12387715646323771032noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-5671748072809412982010-01-12T13:45:40.327+05:302010-01-12T13:45:40.327+05:30രണ്ടു ത്രികോണങ്ങളുടെയും പരപ്പളവുകളുടെ വർഗ്ഗമൂലങ്ങള...രണ്ടു ത്രികോണങ്ങളുടെയും പരപ്പളവുകളുടെ വർഗ്ഗമൂലങ്ങളുടെ തുക ലംബകത്തിന്റെ പരപ്പളവിന്റെ വർഗ്ഗമൂലമായിരിക്കും. വിശദവിവരങ്ങൾ <a href="http://www.usvishakh.net/documents/problems.pdf" rel="nofollow">ഇവിടെ</a> അദ്ധ്യായം 27-ൽ.Umesh::ഉമേഷ്https://www.blogger.com/profile/06474085260183409502noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-64925798324621689352010-01-12T06:52:34.797+05:302010-01-12T06:52:34.797+05:30@ ഉമേഷ് ജീ,
അപാകത ചൂണ്ടിക്കാണിച്ചതിന് നന്ദി.
ഇതാ ത...<b>@ ഉമേഷ് ജീ,<br />അപാകത ചൂണ്ടിക്കാണിച്ചതിന് നന്ദി.<br />ഇതാ തിരുത്തിയിട്ടുണ്ട്.<br /></b>Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-27073395808940230342010-01-12T06:43:24.025+05:302010-01-12T06:43:24.025+05:30ഈ പരപ്പളവും വിസ്തീർണ്ണവും ഒന്നുതന്നെയല്ലേ? എന്തിന...ഈ പരപ്പളവും വിസ്തീർണ്ണവും ഒന്നുതന്നെയല്ലേ? എന്തിനാണു രണ്ടു വാക്കുകൾ ഒരേ പ്രശ്നത്തിൽ ഉപയോഗിച്ചു ചിന്താക്കുഴപ്പം ഉണ്ടാക്കുന്നതു്?<br /><br />Area എന്നു ബ്രായ്ക്കറ്റിൽ കൊടുക്കുന്നതും നന്നായിരിക്കും. ഞാനൊക്കെ പഠിച്ച കാലത്തു നിന്നും തർജ്ജമ പിന്നെയും മാറി.Umesh::ഉമേഷ്https://www.blogger.com/profile/06474085260183409502noreply@blogger.com