Search found 110 matches
- Wed Aug 28, 2013 11:32 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 4
- Replies: 32
- Views: 26802
Re: [OGC1] Online Geometry Camp: Day 4
solution of problem 4: first let the centre of the circle is O. now, $\angle EBF+\angle GCF=180$ $\therefore \frac{1}{2}\angle EBF+\frac{1}{2}\angle GCF=90$ or,$\angle OBF+\angle OCF=90$ or,$\angle BOC=90$ samely we get $\angle BOC=\angle AOD=90$ now let GC=FC=x and GD=HD=y. use pythagorus's theorem...
- Wed Aug 28, 2013 7:04 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 4
- Replies: 32
- Views: 26802
Re: [OGC1] Online Geometry Camp: Day 4
solution of problem 3: let the intersecting point of AD and BE is G. that means G is the centroid of $\Delta ABC$. so,AG=2GD and BG=2GE; now , $AG^{2}+GE^{2}=AE^{2}$ or, $AG^{2}+GE^{2}=(\frac{AC}{2})^{2}$ or, $(2GD)^{2}+GE^{2}=(69.5)^{2}$ or, $4GD^{2}+GE^{2}=4830.25$.....1st result and $BG^{2}+GD^{2...
- Wed Aug 28, 2013 5:34 pm
- Forum: Geometry
- Topic: two circles
- Replies: 3
- Views: 2852
Re: two circles
ohhhh ...i have mitake a line. asif is right. D is the midpoint of AB.
- Wed Aug 28, 2013 12:59 pm
- Forum: Geometry
- Topic: two circles
- Replies: 3
- Views: 2852
two circles
plz give me the solve of this problem..... $\omega_1$ and $ \omega_2$ are two concentric circles where $\omega_1> \omega_2$. A tangent of $\omega_ 2$ at point $B$ touches $\omega_ 1$ at points $A$ and $C$. $D$ is the midpoint of $AB$. Another line through point $A$ touches $\omega_ 2$ at points $E$ ...
- Mon Aug 26, 2013 10:51 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 2
- Replies: 22
- Views: 19305
Re: [OGC1] Online Geometry Camp: Day 2
I have also solved the problem by using stewart's theorem like Samiun Fateeha Ira and i also unable to see any bangla text ....
- Mon Aug 26, 2013 9:04 am
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 1
- Replies: 19
- Views: 17657
Re: [OGC1] Online Geometry Camp: Day 1
$\Delta APQ$ is isoscles and $\angle PAQ$=42. so $\angle PAQ=\angle AQP$=(180-42)/2=69. now compare for $\Delta BDQ$, x+y=$\angle BQA$=$\angle PQA$=$\angle AQP$=69.....1st result and compare for $\Delta BRP$, y+z=180-$\angle BPR$=180-$\angle APQ$=180-69=111....2nd result now multiply 1st and 2nd res...
- Sat Aug 24, 2013 7:46 pm
- Forum: News / Announcements
- Topic: Online Geometry Camp 2014 Phase 1
- Replies: 19
- Views: 15788
Re: Online Geometry Camp 2014 Phase 1
নায়েল ভাইয়ার e-mail adress জানতে চাই...???
- Sat Aug 24, 2013 6:19 pm
- Forum: News / Announcements
- Topic: Online Geometry Camp 2014 Phase 1
- Replies: 19
- Views: 15788
Re: Online Geometry Camp 2014 Phase 1
কুরিয়ার করে পাঠানোর ঠিকানা জানতে চাই। আর উন্মুক্ত আলোচনার কোন সময়ে হবে?......নাকি যেকোন সময়?
- Fri Mar 22, 2013 8:53 pm
- Forum: Geometry
- Topic: Medians and Side Lengths
- Replies: 2
- Views: 2332
Medians and Side Lengths
pls give me the solution of this problem:
Find the ratio of the area of a given tringle to that of a tringle whose sides have the same lengths as the medians of the original triangle.
Find the ratio of the area of a given tringle to that of a tringle whose sides have the same lengths as the medians of the original triangle.
- Thu Mar 21, 2013 3:05 pm
- Forum: Introductions
- Topic: camp info
- Replies: 5
- Views: 6067
camp info
I am new in this forum.....help me......pls inform me the problem of Kaykobad sir