Question
Tue November 08, 2011 By:
 

Prove that the areas of similar triangles have the same ratio as the squares of the radii of their circumcircles.

Expert Reply
Sat November 12, 2011
Let us suppose ?ABC (with sides a,b&c )and ?PQR (with sides p,q&r ) are similar
 
Since these triangles are similar their sides must be in the same ratio
 
That is p/a = q/b = r/c = constant = k
 
Hence p = ak
 
q = bk
 
r = ck
 
Now semi-perimeter of the triangles becomes
for ?ABC, s= a+b+c/2
 
for ?PQR, s' = p+q+r/2 = ka + kb + kc /2 = k(a+b+c)/2 =ks
 
Now the formula for area of a triangle in terms of sides is
for ?ABC, ?1 = ?s(s-a)(s-b)(s-c)    ..........................(1)
 
for ?PQR, ?2 = ?s'(s'-p)(s'-q)(s'-r)
Replace the value of s',p,q&r in terms of s,a,b&c from above we get
 ?2 = ?ks(ks-ka)(ks-kb)(ks-kc)
 
= ?k4s(s-a)(s-b)(s-c) =k2?s(s-a)(s-b)(s-c)  = k2 ?1    (from eqn1)
 
Hence ?2/?1= k2  .......................................... (2)
 
Now the formula for circumradius of a triangle in terms of its sides is given as
for ?ABC , R1 = abc / 4?1
 
for ?PQR , R2 = pqr/ 4?= kakbkc/4?2
Hence from above equations divding Rby R2 we get
 
R1 /R2 = abc / 4?÷ k3abc/4?2
R1 /R2 = abc4?2 / 4?1k3abc = ?2 / ?1k3
Use the value from eqn 2 and we get
R1 /R2 = k2 / k= 1/k
Hence    k = R2 /R1
 
Putting this value of k in eqn 2 we get
 
?2/?1= (R2 /R1)2
 
or, ?1/?2= R12 /R22
 
Proved.
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