# Triangle A has an area of 9  and two sides of lengths 4  and 7 . Triangle B is similar to triangle A and has a side with a length of 16 . What are the maximum and minimum possible areas of triangle B?

Aug 13, 2016

$\textcolor{red}{\text{The maximum possible area of B will be 144}}$

$\textcolor{red}{\text{and the minimum possible area of B will be 47}}$

#### Explanation: Given
$\text{Area Triangle A"=9 " and two sides 4 and 7}$
If the angle between sides 4 & 9 be a then

$\text{Area} = 9 = \frac{1}{2} \cdot 4 \cdot 7 \cdot \sin a$
$\implies a = {\sin}^{-} 1 \left(\frac{9}{14}\right) \approx {40}^{\circ}$

Now if length of the third side be x then

${x}^{2} = {4}^{2} + {7}^{2} - 2 \cdot 4 \cdot 7 \cos {40}^{\circ}$

$x = \sqrt{{4}^{2} + {7}^{2} - 2 \cdot 4 \cdot 7 \cos {40}^{\circ}} \approx 4.7$

So for triangle A

The smallest side has length 4 and largest side has length 7

Now we know that the ratio of areas of two similar triangles is the square of the ratio of their corresponding sides.

${\Delta}_{B} / {\Delta}_{A} = {\left(\text{Length of one side of B"/"Length of Corresponding side of A}\right)}^{2}$

When the side of length 16 of triangle corresponds to the length 4 of triangle A then

${\Delta}_{B} / {\Delta}_{A} = {\left(\frac{16}{4}\right)}^{2}$

$\implies {\Delta}_{B} / 9 = {\left(4\right)}^{2} = 16 \implies {\Delta}_{B} = 9 \times 16 = 144$

Again when the side of length 16 of triangle B corresponds to the length 7 of triangle A then

${\Delta}_{B} / {\Delta}_{A} = {\left(\frac{16}{7}\right)}^{2}$

$\implies {\Delta}_{B} / 9 = \frac{256}{49} = 16 \implies {\Delta}_{B} = 9 \times \frac{256}{49} = 47$

$\textcolor{red}{\text{So the maximum possible area of B will be 144}}$

$\textcolor{red}{\text{and the minimum possible area of B will be 47}}$