Triangle A has an area of 3  and two sides of lengths 3  and 6 . Triangle B is similar to triangle A and has a side with a length of 11 . What are the maximum and minimum possible areas of triangle B?

Feb 17, 2016

The triangle inequality states that the sum of any two sides of a triangle MUST be greater than the 3rd side. That implies the missing side of triangle A must be greater than 3!

Explanation:

Using the triangle inequality ...

$x + 3 > 6$
$x > 3$

So, the missing side of triangle A must fall between 3 and 6.

This means 3 is the shortest side and 6 is the longest side of triangle A.

Since area is proportional to the square of the ratio of the similar sides ...

minimum area $= {\left(\frac{11}{6}\right)}^{2} \times 3 = \frac{121}{12} \approx 10.1$

maximum area $= {\left(\frac{11}{3}\right)}^{2} \times 3 = \frac{121}{3} \approx 40.3$

Hope that helped

P.S. - If you really want to know the length of the missing 3rd side of triangle A, you can use Heron's area formula and determine that the length is $\approx 3.325$. I'll leave that proof to you :)