Finding Missing Side of Similar Triangles

[NK8485]

Hello, I'm working on trying to find the missing side lengths of similar triangles. I understand the Pythagorean theory and proportions but I'm having a tough time when the proportion (k)= a square root.

For instance, this problem I'm working on has two similar right triangles. Upon finding the proportion it is K=(√165)/6

Now to find the missing side of the smaller triangle I need to multiply by 1/K. How do I multiply 1/((√165)/6) by 19, where 19 is the hypotenuse on the larger triangle.


So...

1 (19) = ?
_____
(√165)
_____
6

 

[stapel]

Hello, I'm working on trying to find the missing side lengths of similar triangles. I understand the Pythagorean theory and proportions but I'm having a tough time when the proportion (k)= a square root.

For instance, this problem I'm working on has two similar right triangles. Upon finding the proportion it is K=(√165)/6

Now to find the missing side of the smaller triangle I need to multiply by 1/K. How do I multiply 1/((√165)/6) by 19, where 19 is the hypotenuse on the larger triangle.


So...

1 (19) = ?
_____
(√165)
_____
6

I'm a little lost. What was the original exercise? How did you arrive at the above (triple?) fraction?

 

[Explorer]

Hello, I'm working on trying to find the missing side lengths of similar triangles. I understand the Pythagorean theory and proportions but I'm having a tough time when the proportion (k)= a square root.

For instance, this problem I'm working on has two similar right triangles. Upon finding the proportion it is K=(√165)/6

Now to find the missing side of the smaller triangle I need to multiply by 1/K. How do I multiply 1/((√165)/6) by 19, where 19 is the hypotenuse on the larger triangle.


So...

1 (19) = ?
_____
(√165)
_____
6

Do not try to be fancy by finding the coefficients of proportion K etc. you only get confused.
Use proportions as they should be used. For example, In similar triangles ABC and A'B'C', use something like
AB:A'B' = BC:B'C'

And if you need to find AB:
AB = (A'B' X BC) / B'C'

 

[Ishuda]

Hello, I'm working on trying to find the missing side lengths of similar triangles. I understand the Pythagorean theory and proportions but I'm having a tough time when the proportion (k)= a square root.

For instance, this problem I'm working on has two similar right triangles. Upon finding the proportion it is K=(√165)/6

Now to find the missing side of the smaller triangle I need to multiply by 1/K. How do I multiply 1/((√165)/6) by 19, where 19 is the hypotenuse on the larger triangle.


So...

1 (19) = ?
_____
(√165)
_____
6

Go back to your basic arithmetic: To divide by a fraction, invert and multiply. So if we invert √165/6, we get 6/√165 so we have
$\displaystyle \displaystyle{\frac{1}{\left [\frac{\sqrt{165}}{6}\right ]}\, *\, 19\, =\, \frac{6}{\sqrt{165}}\, *\, 19}$ = ?