# Triangle A has an area of 6  and two sides of lengths 3  and 8 . Triangle B is similar to triangle A and has a side with a length of 7 . What are the maximum and minimum possible areas of triangle B?

Jan 25, 2018

$\text{Area"_"Max} = 10.5$

$\text{Area"_"Min} = 1.43$

#### Explanation:

From the sides and area, we calculate the remaining side of the first triangle as .
$A = \frac{1}{2} b \times h$ with 8 as b, $6 = 4 \times h$; $h = 1.5$
Separating the 8 side into x and 8 - x we obtain:
${3}^{2} = {1.5}^{2} + {x}^{2}$ and ${c}^{2} = {1.5}^{2} + {\left(8 - x\right)}^{2}$
$9 = 2.25 + {x}^{2}$ ; ${x}^{2} = 6.75$; $x = 2.6$
$8 - x = 5.40$
${c}^{2} = {1.5}^{2} + {\left(8 - x\right)}^{2}$; ${c}^{2} = 2.25 + {5.4}^{2}$
$c = 5.60$

The maximum similar triangle 'B' would have 7 as the shortest side, and the minimum would have it as the longest side. Those triangles would have sides of 7, 13.1, 18.7 and 7, 4.90, 2.63 respectively.
The ratio of the base split ($\frac{2.6}{8}$) applied to each gives us:
"Base"_"Max" = 0.325 xx 18.7 = 6.08 (h_"Max")^2 = 7^2 - 6.08^2; h_"Max" = 3.47#

$\text{Base"_"Min} = 2.275$ (derived similar to previous)
${\left({h}_{\text{Min}}\right)}^{2} = {2.6}^{2} - {2.275}^{2}$; ${h}_{\text{Max}} = 1.26$

$\text{Area"_"Max} = \frac{1}{2} \times 6.08 \times 3.47 = 10.5$

$\text{Area"_"Min} = \frac{1}{2} \times 2.275 \times 1.26 = 1.43$