# 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?

Jul 18, 2016

By the square of $\frac{12}{8}$ or the square of $\frac{12}{15}$

#### 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:
$A r e {a}_{\triangle A} = \frac{1}{2} \times 8 \times 15 \sin x = 24$
Hence:
$x = A r c \sin \left(\frac{24}{60}\right)$

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

${L}^{2} = {8}^{2} + {15}^{2} - 2 \times 8 \times 15 \cos x$. 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.

$A r e {a}_{\triangle B} = {r}^{2} \times A r e {a}_{\triangle A}$.

$r$ 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 $r = \frac{12}{8}$. Minimum possible area if $r = \frac{12}{15}$.

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