tag:blogger.com,1999:blog-2787161678592839240.post3206295073715005238..comments2024-03-29T12:30:29.739+05:30Comments on www.mathsblog.in : Maths Blog for School Teachers & Students: സംശയനിവാരണം!വി.കെ. നിസാര്http://www.blogger.com/profile/14303804236214732024noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-2787161678592839240.post-4170785139821516342009-09-20T09:48:39.171+05:302009-09-20T09:48:39.171+05:30Arc BD and Arc AC meet at P
Triangle DPC is eqila...Arc BD and Arc AC meet at P<br />Triangle DPC is eqilateral<br />We cam find area of DPC root 3 10 *10 /4<br />Arc A P has angle 30 .Arc CPB has angle 30<br />Area of these arcs is pie * 10 * 10 *30 / 360<br />Add area of triangle , arce we get unshaded region APB<br />AREa pf square - 4 * area of unshaded APB = 82.4<br /><br />NIMMI AUGUSTINE<br />MATHEMATICS TEACHER<br />ASSISSI VIDYANIKETAN PUBLIC SCHOOL<br />KAKANADAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-16116604594635343512009-09-17T17:28:14.901+05:302009-09-17T17:28:14.901+05:30can we upload any comments with illustration?
(usi...can we upload any comments with illustration?<br />(using diagrams)<br />if so how?VIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-79765826115939964882009-09-17T17:25:11.143+05:302009-09-17T17:25:11.143+05:30my answer is very close to murali sir's answer...my answer is very close to murali sir's answer and similar<br />let the area of the required portion is X.<br />X=100-8*Y (Y=HALF PART OF THE REMAINING PORTION)<br />LET NM be the joining segment of the midpoints of AB& CD and "p" the point of intersection.<br /><br />The area of Y=area of trapeeziam NPDA- area of sector PDA<br />=1/2*5(10+10-5*root3)-pai*10*10*30/360<br />=28.35-26.16<br />=2.19<br />so X=100-8*2.19<br /> =100-17.52<br /> =82.48 SQ CM.VIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-57715527802107246812009-09-17T11:56:07.607+05:302009-09-17T11:56:07.607+05:30dear muraleedharan sir
fine.good qn too
satheesandear muraleedharan sir<br />fine.good qn too<br />satheesanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-10496525216920385322009-09-17T11:25:38.706+05:302009-09-17T11:25:38.706+05:30Dear satheesan sir,
Consider the rightangled trian...Dear satheesan sir,<br />Consider the rightangled triangle KMD<br />KD=Radius=10<br />MD=10/2=5<br />Therefore KM=root of(10*10-5*5)=5*root of 3<br />Since the ratio of sides are 1:root of 3:2, the angles are 30,60,90<br /> <br /> MURALEEDHARAN.C.R,GVHSS VATTENADAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-47907580723474132782009-09-17T07:25:36.005+05:302009-09-17T07:25:36.005+05:30dear muraleeddharan sir
stil some doubts.(1) How a...dear muraleeddharan sir<br />stil some doubts.(1) How angle ADK=30. <br />satheesanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-61063067735355065692009-09-17T05:58:06.681+05:302009-09-17T05:58:06.681+05:30I don't know how a figure is drawn in this spa...I don't know how a figure is drawn in this space<br /><br />Let M be the midpoint of DC & N be the midpoint of AB<br />Join MN<br />Let the point of intrsection of the 2 arcs with centre<br />at D & C be K (it is clear that K is in MN)<br />Required area= area of the square-8*the area of the region bounded by the line segment AN & the arc AK (say Z)<br />ie required area = area of the square -8Z<br />But Z = area of the rectangle ANMD -(area of the sector ADK + area of the triangle KMD)<br /> =50-(3.14*10*10*30/360+1/2*5*5root3<br /> =50-(314/12 +25*1.732/2<br /> =50-(26.167+21.65)<br /> =50-47.817<br /> =2.183<br />ie Z=2.183<br />Required area =100-8*2.183<br /> =100-17.464<br /> =82.536 sq.cm nearly<br />This problem may solve using another method<br />ie Required area = area of the region bounded by the <br />arcs ACs with centres B & D +area of the region bounded by the arcs DBs with centres A & C - area of the common region bouned by the 4 arcs<br />Area of the common region is calculated by using the small square with the same corner points of the common region &4parts of smallregions(small region out side of the small square=sector-triangle)<br /><br /> MURALEEDHARAN.C.R<br /> GVHSS VATTENADAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-68292505506564254192009-09-16T22:23:27.406+05:302009-09-16T22:23:27.406+05:30Yes
I think Lalitha teacher will give a simple , i...Yes<br />I think Lalitha teacher will give a simple , intersting and beautiful answer suitable for our high school students.<br /> I have an answer<br />Area = 2[ 50 pie - 100 ] - K where k is the area of inner portion as we take it twice.<br /><br />How we calculate K?<br />I made 3 equations with three variables x y and z<br />x for middle, y for petal and z for butterfly shape<br />I solved them<br />The answer was nearly 83 by taking usual values of root 3 and pie<br /><br />JOHN P AAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-58293244968923824462009-09-16T21:17:40.892+05:302009-09-16T21:17:40.892+05:30sorry no need of any more time,seems impossible
th...sorry no need of any more time,seems impossible<br />thomasAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-19431474285209839272009-09-16T20:42:31.446+05:302009-09-16T20:42:31.446+05:30give one more day
thomasgive one more day<br />thomasAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-60545088038654256412009-09-16T19:01:45.129+05:302009-09-16T19:01:45.129+05:30you are the person to clear the answer.if any body...you are the person to clear the answer.if any body gives it you may forget it easilyVIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-55890524425633931432009-09-16T18:48:57.081+05:302009-09-16T18:48:57.081+05:30Yes, we are waiting for the answer.
GeethaYes, we are waiting for the answer.<br /><br />GeethaAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-5001820322240124762009-09-16T17:59:12.818+05:302009-09-16T17:59:12.818+05:30for the last two days no one is answering for the ...for the last two days no one is answering for the qn. why? if there is an easy answer it is time to receive it.anyhow the person who sent the qn won in that way.<br />will you accept this answer<br />the area of the shaded portion is " 50pai-100-the area of the centre common portion"vijayanhttps://www.blogger.com/profile/02345885159669791726noreply@blogger.com