# Triangle A has an area of 12  and two sides of lengths 6  and 9 . Triangle B is similar to triangle A and has a side of length 15 . What are the maximum and minimum possible areas of triangle B?

Apr 19, 2017

Maximum area of $\triangle B = 75$

Minimum area of $\triangle B = \frac{100}{3} = 33.3$

#### Explanation:

Similar triangles have identical angles and size ratios. That means the change in length of any side either larger or smaller will be the same for the other two sides. As a result, the area of the $s i m i l a r \triangle ' s$ will also be a ratio of one to the other.

It has been shown that if the ratio of the sides of similar triangles is R, then the ratio of the areas of the triangles is ${R}^{2}$.

Example: For a $3 , 4 , 5 , r i g h t \angle \triangle$ sitting on is $3$ base, its area can be readily calculated form ${A}_{A} = \frac{1}{2} b h = \frac{1}{2} \left(3\right) \left(4\right) = 6$.

But if all three sides are doubled in length, the area of the new triangle is ${A}_{B} = \frac{1}{2} b h = \frac{1}{2} \left(6\right) \left(8\right) = 24$ which is ${2}^{2}$ = 4A_A.

From the information given, we need to find the areas of two new triangles whose sides are increased from either $6 \mathmr{and} 9 \to 15$ that are $s i m i l a r$ to the original two.

Here we have $\triangle A ' s$ with an area $A = 12$ and sides $6 \mathmr{and} 9.$
We also have larger $s i m i l a r \triangle B ' s$ with an area $B$ and side $15.$

The ratio of the change in area of $\triangle A \to \triangle B$ where side $6 \to 15$ is then:

$\triangle B = {\left(\frac{15}{6}\right)}^{2} \triangle A$

$\triangle B = {\left(\frac{15}{6}\right)}^{2} \left(12\right)$

$\triangle B = \left(\frac{225}{\cancel{36} 3}\right) \left(\cancel{12}\right)$

$\triangle B = 75$

The ratio of the change in area of $\triangle A \to \triangle B$ where side $9 \to 15$ is then:

$\triangle B = {\left(\frac{15}{9}\right)}^{2} \triangle A$

$\triangle B = {\left(\frac{15}{9}\right)}^{2} \left(12\right)$

$\triangle B = \left(\frac{225}{\cancel{81} 27}\right) \left(\cancel{12} 4\right)$

$\triangle B = \frac{\cancel{900} 100}{\cancel{27} 3}$

$\triangle B = \frac{100}{3} = 33.3$

Apr 19, 2017

The minimum is $2.567$ and the maximum is $70.772$

#### Explanation:

THIS ANSWER MAY BE INVALID AND IS AWAITING RECALCULATION AND DOUBLE CHECK! Check EET-APs answer for a tried-and-true method of solving the problem.

Because the two triangles are similar, call them triangle $A B C$ and $D E F$, $\frac{A}{D} = \frac{B}{E} = \frac{C}{F}$. We are not given which side has length 15, so we need to calculate it for each value ($A = 6 , B = 9$), and to do this we must find the value of $C$.

Start by recalling Heron's theorem $A = \sqrt{S \left(S - A\right) \left(S - B\right) \left(S - C\right)}$ where $S = \frac{A + B + C}{2}$. $A + B = 15$, so $S = 7.5 + C$. Thus, the equation for the area (substituted for $12$) is 12=sqrt((7.5+C/2)(7.5+C/2-6)(7.5+C/2-9)(7.5+C/2-C). This simplifies to $144 = \left(7.5 + \frac{C}{2}\right) \left(1.5 + \frac{C}{2}\right) \left(7.5 - \frac{C}{2}\right)$, which I will multiply by two for sake of eliminating decimals to get $288 = \left(15 + C\right) \left(3 + C\right) \left(15 - C\right)$. Multiply this out to get $144 = - {C}^{3} - 3 {C}^{2} + 225 C + 675$, $0 = - {C}^{3} - 3 {C}^{2} + 225 C + 531$, $0 = {C}^{3} + 3 {C}^{2} - 225 C - 531$. Factor this to get $C \cong 14.727$.

We can now use this information to find the areas. If $F = 12$, the scale factor between the triangles is $\frac{14.727}{12}$. Multiplying the other two sides by this number yields $D = 13.3635$ and $E \cong 11.045$, and $S \cong 19.568$. Plug this into Heron's formula to get $A = 70.772$. Follow the same set of steps with
$D = 12$ to find that the minimum $A$ approximately equals $2.567$.