# Triangle A has sides of lengths 51 , 45 , and 54 . Triangle B is similar to triangle A and has a side of length 7 . What are the possible lengths of the other two sides of triangle B?

Dec 4, 2017

$\frac{105}{17}$ and $\frac{126}{17}$; or
$\frac{119}{15}$ and $\frac{42}{5}$; or
$\frac{119}{18}$ and $\frac{35}{6}$

#### Explanation:

Two similar triangles have all of their side lengths in the same ratio. So, overall there are 3 possible $\triangle B$s with a length of 7.

Case i) - the 51 length

So lets have the side length 51 go to 7. This is a scale factor of $\frac{7}{51}$. This means we multiply all of the sides by $\frac{7}{51}$

$51 \times \frac{7}{51} = 7$
$45 \times \frac{7}{51} = \frac{315}{51} = \frac{105}{17}$
$54 \times \frac{7}{51} = \frac{126}{17}$

So the lengths are (as fractions) $\frac{105}{17}$ and $\frac{126}{17}$. You can give these as decimals, but generally fractions are better.

Case ii) - the 45 length

We do the same thing here. To get the side of 45 to 7, we multiply by $\frac{7}{45}$

$51 \times \frac{7}{45} = \frac{119}{15}$
$45 \times \frac{7}{45} = 7$
$54 \times \frac{7}{45} = \frac{42}{5}$

So the lengths are $\frac{119}{15}$ and $\frac{42}{5}$

Case iii) - the 54 length

I'm hoping you know what to do by now. We multiply each length by $\frac{7}{54}$

$51 \times \frac{7}{54} = \frac{119}{18}$
$45 \times \frac{7}{54} = \frac{35}{6}$
$54 \times \frac{7}{54} = 7$

So the lengths are $\frac{119}{18}$ and $\frac{35}{6}$

All of these triangles, although they have different side lengths, are all similar to triangle A, and all are answers.