# Triangle A has an area of 6  and two sides of lengths 5  and 3 . Triangle B is similar to triangle A and has a side with a length of 14 . What are the maximum and minimum possible areas of triangle B?

Oct 26, 2017

$\text{Area"_(B"max")=130 2/3" sq.units}$

$\text{Area"_(B"min")=47.04" sq.units}$

#### Explanation:

If $\Delta A$ has an area of $6$ and a base of $3$
then the height of $\Delta A$ (relative to the side with length $3$) is $4$
(Since "Area"_Delta=("base"xx"height")/2)
and
$\Delta A$ is one of the standard right triangles with sides of length $3 , 4 , \mathmr{and} 5$ (see image below if why this is true is not obvious)

If $\Delta B$ has a side of length $14$

• $B$'s maximum area will occur when the side of length $14$ corresponds to $\Delta A$'s side of length $3$
In this case $\Delta B$'s height will be $4 \times \frac{14}{3} = \frac{56}{3}$
and its area will be $\frac{\frac{56}{3} \times 14}{2} = 130 \frac{2}{3}$ (sq. units)

• $B$'s minimum area will occur then the side of length $14$ corresponds to $\Delta A$'s side of length $5$
In this case
$\textcolor{w h i t e}{\text{XXX}} B$'s height will be $4 \times \frac{14}{5} = \frac{56}{5}$
$\textcolor{w h i t e}{\text{XXX}} B$'s base will be $3 \times \frac{14}{5} = \frac{42}{5}$
and
$\textcolor{w h i t e}{\text{XXX}} B$'s area will be $\frac{\frac{56}{5} \times \frac{42}{5}}{2} = \frac{2352}{50} = \frac{4704}{100} = 47.04$ (sq.units)