# Triangle A has an area of 5  and two sides of lengths 9  and 3 . Triangle B is similar to triangle A and has a side with a length of 9 . What are the maximum and minimum possible areas of triangle B?

$45$ & $5$

#### Explanation:

There are two possible cases as follows

Case 1: Let side $9$ of triangle B be the side corresponding to the small side $3$ of triangle A then the ratio of areas $\setminus {\Delta}_{A}$ & $\setminus {\Delta}_{B}$ of similar triangles A & B respectively will be equal to the square of ratio of corresponding sides $3$ & $9$ of both similar triangles hence we have

$\setminus \frac{\setminus {\Delta}_{A}}{\setminus {\Delta}_{B}} = {\left(\frac{3}{9}\right)}^{2}$

$\setminus \frac{5}{\setminus {\Delta}_{B}} = \frac{1}{9} \setminus \quad \left(\setminus \because \setminus {\Delta}_{A} = 5\right)$

$\setminus {\Delta}_{B} = 45$

Case 2: Let side $9$ of triangle B be the side corresponding to the greater side $9$ of triangle A then the ratio of areas $\setminus {\Delta}_{A}$ & $\setminus {\Delta}_{B}$ of similar triangles A & B respectively will be equal to the square of ratio of corresponding sides $9$ & $9$ of both similar triangles hence we have

$\setminus \frac{\setminus {\Delta}_{A}}{\setminus {\Delta}_{B}} = {\left(\frac{9}{9}\right)}^{2}$

$\setminus \frac{5}{\setminus {\Delta}_{B}} = 1 \setminus \quad \left(\setminus \because \setminus {\Delta}_{A} = 5\right)$

$\setminus {\Delta}_{B} = 5$

Hence, maximum possible area of triangle B is $45$ & minimum area is $5$