Question

Triangle A has an area of `6` and two sides of lengths `4` and `6`. Triangle B is similar to triangle A and has a side of length `18`. What are the maximum and minimum possible areas of triangle B?

Answer 1

`A_(BMax) = color(green)(440.8163)`

`A_(BMin) = color(red)(19.8347)`

Explanation:

In Triangle A

p = 4, q = 6. Therefore `(q-p) < r < (q + p)`

i.e. r can have values between 2.1 and 9.9, rounded up to one decimal.

Given triangles A & B are similar

Area of triangle `A_A = 6`

`:. p / x = q / y = r / z` and `hatP = hatX, hatQ = hatY, hatR = hatZ`
`A_A / A_B = ((cancel(1/2)) p r cancel(sin q)) / ((cancel(1/2))x z cancel(sin Y))`

`A_A / A_B = (p/x)^2`

Let side 18 of B proportional to least side 2.1 of A

Then `A_(BMax) = 6 * (18/2.1)^2 = color(green)(440.8163)`

Let side 18 of B proportional to least side 9.9 of A

`A_(BMin) = 6 * (18 / 9.9)^2 = color(red)(19.8347)`