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Two triangles are similar. Triangle $A$ has sides measuring $3$, $4$, and $5$. The shortest side on triangle $B$ measures $6$. What are the lengths of the other two sides of triangle $B$?

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    Try finding a common number than when multiplied by the original sides gives what you need – 2017-01-01
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    What are your thoughts? Do you know what "similar" means in this context? – 2017-01-01

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Let larger side of triangle A is XY = 5, smaller side YZ = 3 and other side XZ = 4.

Now sides of triangle B let larger side PQ = a, smaller QR = 6 and other side PR = b.

Now two triangle are similar so their sides are in proportion.

$\frac{PQ}{XY} = \frac{PR}{XZ} = \frac{QR}{YZ} $

Then you can find other two sides of B.

$\frac{a}{5} = \frac{b}{4} = \frac{6}{3} $

So we have,

$\frac{a}{5} = \frac{6}{3} $

a = 10

Also,

$\frac{b}{4} = \frac{6}{3} $

b = 8

Or you can do like this also,

Smaller side of triangle B is twice than smaller side of triangle A. And similar triangle are in proportion. So you can twice remaining two sides of triangle A to get remaining sides of triangle B.

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Let $x,y$ be respectively the lengths of the medium and longest sides of triangle "$B$". Then similarity of triangles implies: $\dfrac{3}{6} = \dfrac{4}{x} = \dfrac{5}{y}\implies x = 8, y = 10.$

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Okay. So the two triangles are similar, meaning their sides are all proportional, and their angles are congruent. Now the shortest side of triangle A is 3, and the shortest side of triangle B is 6. Simple division shows that triangle B is exactly twice as big as A. Therefore using 2 as a scale factor, the remaining sides are 8 and 10.