# Triangle A has an area of 15  and two sides of lengths 5  and 9 . Triangle B is similar to triangle A and has a side of length 12 . What are the maximum and minimum possible areas of triangle B?

Dec 21, 2017

Maximum possible area of triangle A = $\textcolor{g r e e n}{128.4949}$

Minimum possible area of triangle B = $\textcolor{red}{11.1795}$

#### Explanation:

$\Delta s A \mathmr{and} B$ are similar.

To get the maximum area of $\Delta B$, side 12 of $\Delta B$ should correspond to side $\left(> 9 - 5\right)$ of $\Delta A$ say $\textcolor{red}{4.1}$ as sum of two sides must be greater than the third side of the triangle (corrected to one decimal point)

Sides are in the ratio 12 : 4.1
Hence the areas will be in the ratio of ${12}^{2} : {\left(4.1\right)}^{2}$

Maximum Area of triangle $B = 15 \cdot {\left(\frac{12}{4.1}\right)}^{2} = \textcolor{g r e e n}{128.4949}$

Similarly to get the minimum area, side 12 of $\Delta B$ will correspond to side <9 + 5) of $\Delta A$. Say $\textcolor{g r e e n}{13.9}$ as sum of two sides must be greater than the third side of the triangle (corrected to one decimal point)
Sides are in the ratio $12 : 13.9$ and areas ${12}^{2} : {13.9}^{2}$

Minimum area of $\Delta B = 15 \cdot {\left(\frac{12}{13.9}\right)}^{2} = \textcolor{red}{11.1795}$

Dec 21, 2017

Maximum Area of ${\triangle}_{B} = 60$ sq. units
Minimum Area of ${\triangle}_{B} \approx 13.6$ sq. units

#### Explanation:

If ${\triangle}_{A}$ has two sides $a = 7$ and $b = 8$ and an area ${\text{Area}}_{A} = 15$
then the length of the third side $c$ can (through manipulating Heron's formula) be derived as:
$\textcolor{w h i t e}{{\text{XXX")c^2=a^2+b^2+-2sqrt(a^2b^2-4"Area}}_{A}}$

Using a calculator we find two possible values for $c$
$c \approx 9.65 \textcolor{w h i t e}{\text{xxx)orcolor(white)("xxx}} c \approx 14.70$

If two triangles ${\triangle}_{A}$ and ${\triangle}_{B}$ are similar then their area vary as the square of corresponding side lengths:
That is
color(white)("XXX")"Area"_B="Area"_A * (("side"_B)/("side"_A))^2

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Given ${\text{Area}}_{A} = 15$ and ${\text{side}}_{B} = 14$
then ${\text{Area}}_{B}$ will be a maximum when the ratio $\left({\text{side"_B)/("side}}_{A}\right)$ is a maximum;
that is when ${\text{side}}_{B}$ corresponds to the minimum possible corresponding value for $s i {\mathrm{de}}_{A}$, namely $7$

${\text{Area}}_{B}$ will be a maximum $15 \cdot {\left(\frac{14}{7}\right)}^{2} = 60$

~~~~~~~~~~~~~~~~~~~~~

Given ${\text{Area}}_{A} = 15$ and ${\text{side}}_{B} = 14$
then ${\text{Area}}_{B}$ will be a minimum when the ratio $\left({\text{side"_B)/("side}}_{A}\right)$ is a minimum;
that is when ${\text{side}}_{B}$ corresponds to the maximum possible corresponding value for $s i {\mathrm{de}}_{A}$, namely $14.70$ (based on our earlier analysis)

${\text{Area}}_{B}$ will be a minimum $15 \cdot {\left(\frac{14}{14.7}\right)}^{2} \approx 13.60$