Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Let ABCD be a tetrahedron with AB=41,BC=36,CA=7,DA=18,DB=27,DC=13.Find the distance between the mid points of edges AB and CD ?
Dear sindhuja
In any triangle ABC having sides a,b,c Median AD is given as
AD2 = 1/4 ( 2b2 +2c2-a2)
so
let mid point of AB is E and mid point CD is F
in triangle ABC
CE2 = 1/4( 2*362 + 2*72 -412)
in triangle ADB
DE2 = 1/4( 2*272 + 2*182 -412)
and in triangle CDE
EF2 = 1/4( 2*CE2 +2*DE2 - 132)
put the value
EF =√137
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !