# Triangle A has an area of 12  and two sides of lengths 4  and 8 . Triangle B is similar to triangle A and has a side of length 7 . What are the maximum and minimum possible areas of triangle B?

Oct 23, 2017

${A}_{\text{Bmin}} \approx 4.8$
${A}_{\text{Bmax}} = 36.75$

#### Explanation:

First you must find the side lengths for the maximum sized triangle A , when the longest side is greater than 4 and 8 and the minimum sized triangle , when 8 is the longest side.

To do this use Heron's Area formula : $s = \frac{a + b + c}{2}$ where a, b, & c are the side lengths of the triangle:

$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

Let $a = 8 , b = 4 \text{ & "c " is unknown side lengths}$

$s = \frac{12 + c}{2} = 6 + \frac{1}{2} c$

${A}_{A} = 12 = \sqrt{\left(6 + \frac{1}{2} c\right) \left(6 + \frac{1}{2} c - 4\right) \left(6 + \frac{1}{2} c - 8\right) \left(6 + \frac{1}{2} c - c\right)}$

${A}_{A} = 12 = \sqrt{\left(6 + \frac{1}{2} c\right) \left(2 + \frac{1}{2} c\right) \left(- 2 + \frac{1}{2} c\right) \left(6 - \frac{1}{2} c\right)}$

Square both sides:

$144 = \left(6 + \frac{1}{2} c\right) \left(2 + \frac{1}{2} c\right) \left(- 2 + \frac{1}{2} c\right) \left(6 - \frac{1}{2} c\right)$

Pull out a 1/2 from each factor:

$144 = \frac{1}{16} \left(12 + c\right) \left(4 + c\right) \left(- 4 + c\right) \left(12 - c\right)$

Simplify:

$2304 = \left(12 + c\right) \left(4 + c\right) \left(- 4 + c\right) \left(12 - c\right)$

$2304 = \left(48 + 8 c - {c}^{2}\right) \left(- 48 + 8 c + {c}^{2}\right)$

$2304 = - 2304 + 384 c + 48 {c}^{2} - 384 c + 64 {c}^{2} + 8 {c}^{3} + 48 {c}^{2} - 8 {c}^{3} - {c}^{4}$

${c}^{4} - 160 {c}^{2} + 4608 = 0$

*Substitute $x = {c}^{2} \cdot : \text{ } {x}^{2} - 160 x + 4608 = 0$

Use completing the square:

$\left({x}^{2} - 160 x\right) = - 4608$

${\left(x - \frac{160}{2}\right)}^{2} = - 4608 + {\left(- \frac{160}{2}\right)}^{2}$

${\left(x - 80\right)}^{2} = 1792$

Square root both sides:

$x - 80 = \pm \sqrt{1792}$

$x = 80 \pm \sqrt{16} \sqrt{16} \sqrt{7}$

$x = 80 \pm 16 \sqrt{7}$

Substitute ${c}^{2} = x$:

${c}^{2} = 80 \pm 16 \sqrt{7}$

$c = \pm \sqrt{80 \pm 16 \sqrt{7}}$

Since triangle side lengths are positive we need to ignore the negative answers:

Minimum and maximum side lengths of triangle A:
$c = \sqrt{80 \pm 16 \sqrt{7}} \approx 6.137 , 11.06$

Since the area of triangles are proportional to the square of the side lengths we can find the maximum and minimum areas of triangle B:

A_B/A_A = (7/4)^2; " "A_B = (7/4)^2 * 12 = 36.75

A_B/A_A = (7/8)^2; " "A_B = (7/8)^2 * 12 = 9.1875

A_B/A_A ~~ (7/11.06)^2; " "A_B ~~ (7/11.06)^2 * 12 ~~ 4.8

A_B/A_A ~~ (7/6.137)^2; " "A_B ~~ (7/6.137)^2 * 12 ~~15.6

${A}_{\text{Bmin}} \approx 4.8$
${A}_{\text{Bmax}} = 36.75$