# Triangle A has an area of 9  and two sides of lengths 6  and 9 . Triangle B is similar to triangle A and has a side of length 12 . What are the maximum and minimum possible areas of triangle B?

Jun 29, 2018

Min $= \setminus \frac{144 \left(13 - 8 \setminus \sqrt{2}\right)}{41} \setminus \approx 5.922584784 \ldots$
Max $= \setminus \frac{144 \left(13 + 8 \setminus \sqrt{2}\right)}{41} \setminus \approx 85.39448839 \ldots$

#### Explanation:

Given:
$A r e {a}_{\setminus \triangle A} = 9$
Side lengths of $\setminus \triangle A$ are $X , Y , Z$
$X = 6 , Y = 9$
Side lengths of $\setminus \triangle B$ are $U , V , W$
$U = 12$
$\setminus \triangle A \setminus \setminus \textrm{s i m i l a r} \setminus \triangle B$

first solve for $Z$:
use Heron's Formula:  A = \sqrt{S(S-A)(S-B)(S-C)  where $S = \setminus \frac{A + B + C}{2}$ , sub in area 9, and sidelengths 6 and 9 .

$S = \setminus \frac{15 + z}{2}$
 9 = \sqrt{(\frac{15 + Z}{2})(\frac{Z+ 3}{2})(\frac{Z - 3}{2})(\frac{15 - z}{2})
$81 = \setminus \frac{\left(225 - {Z}^{2}\right) \left({Z}^{2} - 9\right)}{16}$
$1296 = - {Z}^{4} + 234 {Z}^{2} - 2025$
$- {Z}^{4} + 234 {Z}^{2} - 3321 = 0$

Let $u = {Z}^{2}$, $- {u}^{2} + 234 u - 3321 = 0$
$u = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$u = 9 \left(13 - 8 \setminus \sqrt{2}\right) , u = 9 \left(8 \setminus \sqrt{2} + 13\right)$

$Z = \setminus \sqrt{u}$ Reject the negative solutions as $Z > 0$
$Z = 3 \setminus \sqrt{13 - 8 \setminus \sqrt{2}} , Z = 3 \setminus \sqrt{8 \setminus \sqrt{2} + 13}$
Thus $Z \setminus \approx 3.895718613$ and $14.79267983$ respectively

$\setminus \because \setminus \triangle A \setminus \setminus \textrm{s i m i l a r} \setminus \triangle B , A r e {a}_{\setminus \triangle B} = {k}^{2} \cdot A r e {a}_{\setminus \triangle A}$ where $k$ is the the resizing factor

$k = \frac{12}{s}$ where arranged in ascending order: $s \setminus \in \left\{3 \setminus \sqrt{13 - 8 \setminus \sqrt{2}} , 6 , 9 , 3 \setminus \sqrt{8 \setminus \sqrt{2} + 13}\right\}$
or in decimal form: $s \setminus \in \left\{3.895718613 , 6 , 9 , 14.79267983\right\}$

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 $s = 3 \setminus \sqrt{13 - 8 \setminus \sqrt{2}}$
and to maximize Area choose $s = 3 \setminus \sqrt{8 \setminus \sqrt{2} + 13}$

Thus, minimum Area $= 9 \cdot {\left[\setminus \frac{12}{3 \setminus \sqrt{8 \setminus \sqrt{2} + 13}}\right]}^{2}$
$= \setminus \frac{144 \left(13 - 8 \setminus \sqrt{2}\right)}{41} \setminus \approx 5.922584784 \ldots$

and the maximum Area $= 9 \cdot {\left[\setminus \frac{12}{3 \setminus \sqrt{13 - 8 \setminus \sqrt{2}}}\right]}^{2}$
$= \setminus \frac{144 \left(13 + 8 \setminus \sqrt{2}\right)}{41} \setminus \approx 85.39448839 \ldots$