1any conformations to the above .. or any shorter method?
show that the least perimeter of an isosceles triangle in which a circle of radius 'r' can be inscribed is 6√3r...????
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2 Answers
1No answers to this one??
let the sides be a,a,b
then
2a+b=P=2s where s is the semiperimeter --- (i)
differentiating both sides w.r.t b
2da/db + 1=2ds/db=dP/db ------- (ii)
area of the triangle=rs where r given is the inradius
also area=√(s(s-a)(s-a)(s-b))
so
r2s2=s(s-a)2(s-b) ------(iii)
hence as we know a as a function of b s will come as a function of b too
at maximum perimeter
dP/db=0
hence d(2s)/db=0
ds/db=0 ---------(v)
thus from (ii)
da/db=-1/2
further from (iii)
differentiating it w.r.t b
d/db(r2s)=d/db((s-a)2(s-b))
at maximum perimeter ds/db=0
hence we get
2(s-a)(s-b)(ds/db-da/db)+ (s-a)2(ds/db-1)=0
from(v)
da/db=-1/2
hence
2(s-a)(s-b)(1/2) + (s-a)2(-1)=0
we get
s-a=s-b
or a=b
thus maximum perimeter in case of an equilateral triangle ..
side of triangle=
2(r/tan30)=2√3r
perimeter= side*3=6√3r
