Question

Triangle A has an area of `24` and two sides of lengths `8` and `15`. Triangle B is similar to triangle A and has a side with a length of `12`. What are the maximum and minimum possible areas of triangle B?

Answer 1

By the square of `12/8` or the square of `12/15`

Explanation:

We know that triangle A has fixed internal angles with the given information. Right now we are only interested in the angle between lengths `8&15`.

That angle is in the relationship:
`Area_(triangle A)=1/2xx8xx15sinx=24`
Hence:
`x=Arcsin(24/60)`

With that angle, we can now find the length of the third arm of `triangle A` using the cosine rule .

`L^2=8^2+15^2-2xx8xx15cosx`. Since `x` is already known,
`L=8.3`.

From `triangle A`, we now know for sure that the longest and shortest arms are 15 and 8 respectively.

Similar triangles will have their ratios of arms extended or contracted by a fixed ratio. If one arm doubles in length, the other arms double as well . For area of a similar triangle, if the length of arms double, the area is a size bigger by a factor of 4.

`Area_(triangle B)=r^2xxArea_(triangle A)`.

`r` is the ratio of any side of B to the same side of A.

A similar `triangle B` with an unspecified side 12 will have a maximum area if the ratio is the largest possible hence `r=12/8`. Minimum possible area if `r=12/15`.

Therefore maximum area of B is 54 and the minimum area is 15.36.