# Triangle A has an area of 18  and two sides of lengths 9  and 6 . Triangle B is similar to triangle A and has a side of length 8 . What are the maximum and minimum possible areas of triangle B?

Nov 8, 2016

$32$ and $\frac{128}{9}$

#### Explanation:

Let $\alpha$ be the angle between sides 9 and 6
$A = \frac{9 \cdot 6}{2} \cdot \sin \alpha$

$\sin \alpha = \frac{18 \cdot 2}{9 \cdot 6} = \frac{2}{3} \setminus \setminus \implies \cos \alpha = \frac{\sqrt{5}}{3}$

Let calculate third side $c$ using law of cosines

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \alpha = 81 + 36 - 2 \cdot 9 \cdot 6 \cdot \frac{\sqrt{5}}{3} = 117 - 36 \sqrt{5}$

$c = \sqrt{117 - 36 \sqrt{5}} \cong 6.04$

So 9 is the longest side and 6 is the shortest, so the maximum area is obtained when 8 is proportional to 6 and the minimum when 8 is proportional to 9:

${A}_{M A X} = {\left(\frac{8}{6}\right)}^{2} \cdot 18 = \frac{16}{9} \cdot 18 = 32$

${A}_{M I N} = {\left(\frac{8}{9}\right)}^{2} \cdot 18 = \frac{64}{81} \cdot 18 = \frac{128}{9}$