Question

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?

Answer 1

`A_"Bmin"~~ 4.8`
`A_"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 = (a+b+c)/2` where `a, b, & c` are the side lengths of the triangle:

`A = sqrt(s(s-a)(s-b)(s-c))`

Let `a = 8, b = 4 " & "c " is unknown side lengths"`

`s = (12+c)/2 = 6 + 1/2c`

`A_A = 12 = sqrt((6 + 1/2c)(6 + 1/2c-4)(6 + 1/2c-8)(6 + 1/2c-c))`

`A_A = 12 = sqrt((6 + 1/2c)(2+1/2c)(-2+1/2c)(6-1/2c))`

Square both sides:

`144 = (6 + 1/2c)(2+1/2c)(-2+1/2c)(6-1/2c)`

Pull out a 1/2 from each factor:

`144 = 1/16(12+c)(4+c)(-4+c)(12-c)`

Simplify:

`2304 = (12+c)(4+c)(-4+c)(12-c)`

`2304 = (48+8c-c^2)(-48+8c+c^2)`

`2304 = -2304 +384c + 48c^2 - 384c + 64c^2 + 8c^3+48c^2-8c^3-c^4`

`c^4 - 160c^2 + 4608 = 0`

*Substitute `x = c^2* : " "x^2 -160x + 4608 = 0`

Use completing the square:

`(x^2-160x) = -4608`

`(x - 160/2)^2 = -4608 + (-160/2)^2`

`(x-80)^2 = 1792`

Square root both sides:

`x-80 = +-sqrt(1792)`

`x = 80 +-sqrt(16)sqrt(16)sqrt(7)`

`x = 80 +-16 sqrt(7)`

Substitute `c^2 = x`:

`c^2 = 80 +-16 sqrt(7)`

`c = +- sqrt(80 +-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 +-16 sqrt(7)) ~~ 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_"Bmin"~~ 4.8`
`A_"Bmax"= 36.75`