# Triangle A has an area of 6  and two sides of lengths 4  and 6 . Triangle B is similar to triangle A and has a side of length 18 . What are the maximum and minimum possible areas of triangle B?

Feb 4, 2018

${A}_{B M a x} = \textcolor{g r e e n}{440.8163}$

${A}_{B M \in} = \textcolor{red}{19.8347}$

#### Explanation: In Triangle A

p = 4, q = 6. Therefore $\left(q - p\right) < r < \left(q + p\right)$

i.e. r can have values between 2.1 and 9.9, rounded up to one decimal.

Given triangles A & B are similar

Area of triangle ${A}_{A} = 6$

$\therefore \frac{p}{x} = \frac{q}{y} = \frac{r}{z}$ and $\hat{P} = \hat{X} , \hat{Q} = \hat{Y} , \hat{R} = \hat{Z}$
${A}_{A} / {A}_{B} = \frac{\left(\cancel{\frac{1}{2}}\right) p r \cancel{\sin q}}{\left(\cancel{\frac{1}{2}}\right) x z \cancel{\sin Y}}$

${A}_{A} / {A}_{B} = {\left(\frac{p}{x}\right)}^{2}$

Let side 18 of B proportional to least side 2.1 of A

Then ${A}_{B M a x} = 6 \cdot {\left(\frac{18}{2.1}\right)}^{2} = \textcolor{g r e e n}{440.8163}$

Let side 18 of B proportional to least side 9.9 of A

${A}_{B M \in} = 6 \cdot {\left(\frac{18}{9.9}\right)}^{2} = \textcolor{red}{19.8347}$