21@debo i din thought of appolonius theorem though....perhaps in the soln booklet they must hav given soln with tat only....but how does it matter...i did it in another way....though slight lengthy[1][3]
In a triangle with BC = 3 & AC = 4 & circle with AB as diameter passes thru the centroid of a
triangle then AB is
(A) √5
(B) 5
(C) √3
(D) None
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7 Answers
21see its easy
let vertex A be origin(0,0)
vertex B be (k,0)
vertex C be(p,q)
now eq of circle with AB as diameter is
(x-k2)2 + y2=k2/4......................i
now centriod of triangle is (p+k3 ,q3)
since circle passe through it we get
(p+k3-k2)2 +(q3)2 =k2/4 ................ii
now since we know AC is 4 and BC is 3
so we get p2+q2=16...........................ii
(p-k)2 +q2=9......................................iii
now we hav 3 eq and 3 variables
eliminating p and q we get k =√5.....so length of side AB IS √5
19no need to take such calculations
....use apollonius theorem ! its done!
1how wud u use appolonius theorem here....i tried but cudnt get debtosh....cud u plz show
1CD/3 = r => CD=3r
AC2 + BC2 = 2( CD2+ BD2)
=> 25 = 2( r2+9r2)
=> 25 = 2(10 r2)
=> 2r = √5
21
19we must not go for lengthy , customary methods in these exams....it takes away a lot of time,,,,just thinking for 1 minute can give you a very brief method ,,,just as i i got while doing this sum !