Given:
Area_{\triangleA} = 9
Side lengths of \triangleA are X,Y,Z
X = 6, Y = 9
Side lengths of \triangleB are U,V,W
U = 12
\triangle A \ \text{similar} \triangle B
first solve for Z:
use Heron's Formula: A = \sqrt{S(S-A)(S-B)(S-C) where S = \frac{A+B+C}{2} , sub in area 9, and sidelengths 6 and 9 .
S = \frac{15 + z}{2}
9 = \sqrt{(\frac{15 + Z}{2})(\frac{Z+ 3}{2})(\frac{Z - 3}{2})(\frac{15 - z}{2})
81 = \frac{(225-Z^2)(Z^2 - 9)}{16}
1296 = -Z^4+234Z^2-2025
-Z^4+234Z^2-3321= 0
Let u = Z^2 , -u^2+234u-3321=0
use quadratic formula
u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}
u=9(13-8\sqrt{2}), u=9(8\sqrt{2}+13)
Z = \sqrt{u} Reject the negative solutions as Z>0
Z=3\sqrt{13-8\sqrt{2}}, Z=3\sqrt{8\sqrt{2}+13}
Thus Z \approx 3.895718613 and 14.79267983 respectively
\because \triangle A \ \text{similar} \triangle B, Area_{\triangle B} = k^2 * Area_{\triangleA} where k is the the resizing factor
k = 12/s where arranged in ascending order: s \in { 3\sqrt{13-8\sqrt{2}}, 6, 9,3\sqrt{8\sqrt{2}+13}}
or in decimal form: s \in { 3.895718613, 6, 9,14.79267983}
The greater the value of s , the smaller the Area and the smaller the value of s , the greater the Area,
Thus, to minimize Area choose s = 3\sqrt{13-8\sqrt{2}}
and to maximize Area choose s = 3\sqrt{8\sqrt{2}+13}
Thus, minimum Area = 9 * [ \frac{12}{3\sqrt{8\sqrt{2}+13}}]^2
= \frac{144(13 -8\sqrt{2})}{41} \approx 5.922584784...
and the maximum Area = 9 * [ \frac{12}{3\sqrt{13-8\sqrt{2}}}]^2
= \frac{144(13+8\sqrt{2})}{41} \approx 85.39448839...
