# Triangle A has an area of 4  and two sides of lengths 8  and 4 . Triangle B is similar to triangle A and has a side with a length of 13 . What are the maximum and minimum possible areas of triangle B?

Mar 31, 2017

$\text{Max} = \frac{169}{40} \left(5 + \sqrt{15}\right) \approx 37.488$
$\text{Min} = \frac{169}{40} \left(5 - \sqrt{15}\right) \approx 4.762$

#### Explanation:

Let the vertices of triangle $A$ be labelled $P$, $Q$, $R$, with $P Q = 8$ and $Q R = 4$.

Using Heron's Formula,

$\text{Area} = \sqrt{S \left(S - P Q\right) \left(S - Q R\right) \left(S - P R\right)}$, where

$S = \frac{P Q + Q R + P R}{2}$ is the half-perimeter,

we have

$S = \frac{8 + 4 + P R}{2} = \frac{12 + P R}{2}$

Thus,

$\sqrt{S \left(S - P Q\right) \left(S - Q R\right) \left(S - P R\right)}$

$= \sqrt{\left(\frac{12 + P Q}{2}\right) \left(\frac{12 + P Q}{2} - 8\right) \left(\frac{12 + P Q}{2} - 4\right) \left(\frac{12 + P Q}{2} - P Q\right)}$

$= \frac{\sqrt{\left(12 + P Q\right) \left(P Q - 4\right) \left(4 + P Q\right) \left(12 - P Q\right)}}{4}$

$= \text{Area} = 4$

Solve for $C$.

$\sqrt{\left(144 - P {Q}^{2}\right) \left(P {Q}^{2} - 16\right)} = 16$

$\left(P {Q}^{2} - 144\right) \left(P {Q}^{2} - 16\right) = - 256$

$P {Q}^{4} - 160 P {Q}^{2} + 2304 = - 256$

${\left(P {Q}^{2}\right)}^{2} - 160 P {Q}^{2} + 2560 = 0$

Complete the square.

${\left({\left(P {Q}^{2}\right)}^{2} - 80\right)}^{2} + 2560 = {80}^{2}$

${\left({\left(P {Q}^{2}\right)}^{2} - 80\right)}^{2} = 3840$

$P {Q}^{2} = 80 + 16 \sqrt{15}$ or $P {Q}^{2} = 80 - 16 \sqrt{15}$

$P Q = 4 \sqrt{5 + \sqrt{15}} \approx 11.915$ or
$P Q = 4 \sqrt{5 - \sqrt{15}} \approx 4.246$

This shows that there are 2 possible kinds of triangle that satisfy the conditions given.

In the case of max area for triangle be, we want the side with length 13 to be similar to the side PQ for the triangle with $P Q = 4 \sqrt{5 - \sqrt{15}} \approx 4.246$.

Therefore, the linear scale ratio is

$\frac{13}{4 \sqrt{5 - \sqrt{15}}} \approx 3.061$

The area is therefore enlarged to a factor that is the square of the linear scale ratio. Therefore, The max area triangle B can have is

$4 \times {\left(\frac{13}{4 \sqrt{5 - \sqrt{15}}}\right)}^{2} = \frac{169}{40} \left(5 + \sqrt{15}\right) \approx 37.488$

Similarly, in the case of min area for triangle be, we want the side with length 13 to be similar to the side PQ for the triangle with $P Q = 4 \sqrt{5 + \sqrt{15}} \approx 11.915$.

Therefore, the linear scale ratio is

$\frac{13}{4 \sqrt{5 + \sqrt{15}}} \approx 1.091$

The area is therefore enlarged to a factor that is the square of the linear scale ratio. Therefore, The min area triangle B can have is

$4 \times {\left(\frac{13}{4 \sqrt{5 + \sqrt{15}}}\right)}^{2} = \frac{169}{40} \left(5 - \sqrt{15}\right) \approx 4.762$