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...#