aits pt2

see 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

no need to take such calculations
....use apollonius theorem ! its done!

how wud u use appolonius theorem here....i tried but cudnt get debtosh....cud u plz show

CD/3 = r => CD=3r

AC² + BC² = 2( CD²+ BD²)

=> 25 = 2( r²+9r²)

=> 25 = 2(10 r²)

=> 2r = √5

@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]

we 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 !