# Triangle A has an area of 4  and two sides of lengths 6  and 4 . 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?

Jan 23, 2018

${A}_{\min} = \textcolor{red}{3.3058}$

${A}_{\max} = \textcolor{g r e e n}{73.4694}$

#### Explanation:

Let the areas of triangles be A1 & A2 and sides a1 & a2.

Condition for the triangle’s third side: Sum of the two sides must be greater than the third side.

In our case the given two sides are 6, 4.

Third side should be less than 10 and greater than 2.

Hence the third side will have the maximum value 9.9 and the minimum value 2.1. (Corrected upto one decimal point)

Areas will be proportional to the (side)^2.

$A 2 = A 1 \cdot \left(\frac{a 2}{a 1} ^ 2\right)$

Case : Minimum Area :

When the similar triangle’s side 9 corresponds to 9.9, we get he Minimum area of the triangle.

${A}_{\min} = 4 \cdot {\left(\frac{9}{9.9}\right)}^{2} = \textcolor{red}{3.3058}$

Case : Maximum Area :

When the similar triangle’s side 9 corresponds to 2.1, we get he Maximum area of the triangle.

${A}_{\max} = 4 \cdot {\left(\frac{9}{2.1}\right)}^{2} = \textcolor{g r e e n}{73.4694}$