# Triangle A has an area of 15  and two sides of lengths 8  and 7 . Triangle B is similar to triangle A and has a side with a length of 14 . What are the maximum and minimum possible areas of triangle B?

Dec 21, 2017

Maximum possible area of triangle B = 60
Minimum possible area of triangle B = 45.9375

#### Explanation:

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

To get the maximum area of $\Delta B$, side 14 of $\Delta B$ should correspond to side 7 of $\Delta A$.

Sides are in the ratio 14 : 7
Hence the areas will be in the ratio of ${14}^{2} : {7}^{2} = 196 : 49$

Maximum Area of triangle $B = \frac{15 \cdot 196}{49} = 60$

Similarly to get the minimum area, side 8 of $\Delta A$ will correspond to side 14 of $\Delta B$.
Sides are in the ratio $14 : 8$ and areas $196 : 64$

Minimum area of $\Delta B = \frac{15 \cdot 196}{64} = 45.9375$

Dec 21, 2017

Maximum area: $\approx 159.5$ sq. units
Minimum area: $\approx 14.2$ sq. units

#### Explanation:

If ${\triangle}_{A}$ has sides $a = 7$, $b = 8$, c=? and an area of $A = 15$
then $c \approx 4.3 \textcolor{w h i t e}{\text{XXX")"or"color(white)("XXX}} c \approx 14.4$
(See below for indication of how these values were derived).

Therefore $\triangle A$ could have a minimum side length of $4.3$ (approx)
and a maximum side length of $14.4$ (approx.)

For corresponding sides:
color(white)("XXX")("Area"_B)/("Area"_A)=(("Side"_B)/("Side"_A))^2
or equivalently

color(white)("XXX")"Area"_B="Area"_A * (("Side"_B)/("Side"_A))^2

Notice that the greater the length of the corresponding ${\text{Side}}_{A}$,
the smaller the value of ${\text{Area}}_{B}$
So given ${\text{Area}}_{A} = 15$
and ${\text{Side}}_{B} = 14$
and the maximum value for a corresponding side is ${\text{Side}}_{A} \approx 14.4$
the minimum area for $\triangle B$ is $15 \cdot {\left(\frac{14}{14.4}\right)}^{2} \approx 14.164$

Similarly, notice that the smalle the length of the corresponding ${\text{Side}}_{A}$,
the greater the value of ${\text{Area}}_{B}$
So given ${\text{Area}}_{A} = 15$
and ${\text{Side}}_{B} = 14$
and the minimum value for a corresponding side is ${\text{Side}}_{A} \approx 4.3$
the maximum area for $\triangle B$ is 15 * (14/4.3)^2 ~~159.546

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

Determining possible lengths for $c$

Suppose we place $\triangle A$ on a standard Cartesian plane with the side with length $8$ along the positive X-axis from $x = 0$ to $x = 8$
Using this side as a base and given that the Area of $\triangle A$ is $15$
we see that the vertex opposite this side must be at a height of $y = \frac{15}{4}$

If the side with length $7$ has one end at the origin (coterminal there with the side of length 8) then the other end of the side with length $7$ must be on the circle ${x}^{2} + {y}^{2} = {7}^{2}$
(Note that the other end of the line of length $7$ must be the vertex opposite the side with length $8$)

Substituting, we have
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(\frac{15}{4}\right)}^{2} = {7}^{2}$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 559 ' 16$

$\textcolor{w h i t e}{\text{XXX}} x = \pm \frac{\sqrt{559}}{4}$

Giving possible coordinates: $\left(- \frac{\sqrt{559}}{4} , \frac{15}{4}\right)$ and $\left(+ \frac{\sqrt{559}}{4} , \frac{15}{4}\right)$

We can then use the Pythagorean Theorem to calculate the distance to each of the points from $\left(8 , 0\right)$
giving the possible values shown above (Sorry, details missing but Socratic is already complaining about the length).