# Triangle A has an area of 6  and two sides of lengths 5  and 8 . Triangle B is similar to triangle A and has a side with a length of 19 . What are the maximum and minimum possible areas of triangle B?

##### 1 Answer
Feb 2, 2018

Case 1. Maximum Area of $\Delta B$ = $\textcolor{g r e e n}{225.3902}$ sq. units

Case 2. Minimum Area of $\Delta B$ = $\textcolor{red}{13.016}$ sq. units

#### Explanation:

$\Delta A = 6 , p = 5 , q = 8 , x = 19$

Side r can have values between q-p) , (q + p)

$r > \left(8 - 5\right) , < \left(8 + 5\right)$

$r > 3 < 13$

${r}_{\min} = 3.1 , {r}_{\max} = 12.9$, rounded to one decemal.

Case 1. Maximum Area of $\Delta B$

x should correspond to least side of A, viz. r = 3.1 to get minimum area of B.

$\Delta \frac{B}{\Delta} A = {\left(\frac{x}{r}\right)}^{2}$

$\Delta B = 6 \cdot {\left(\frac{19}{3.1}\right)}^{2} = \textcolor{g r e e n}{225.3902}$ sq. units

Case 2. Minimum Area of $\Delta B$

x should correspond to longest side of A, viz. r = 12.9 to get maximum area of B.

$\Delta \frac{B}{\Delta} A = {\left(\frac{x}{r}\right)}^{2}$

$\Delta B = 6 \cdot {\left(\frac{19}{12.9}\right)}^{2} = \textcolor{red}{13.016}$ sq. units