Question

12/16=y/x
what are the values of x and y?

diagram below

Answer 1

do you want the choices?
@mathmate

Answer 2

1. The part with length 16 is shorter than the part marked 12. We have to assume that the diagram is not to scale.
2. By metric relations of right triangles,
\(x^2=16*(16+12)= 448\) which means that \(x=8\sqrt7\)
\(y^2=12*(16+12)= 336\) which means that \(y=4\sqrt{21}\)
So \(y/x=\dfrac{4\sqrt{21}}{8\sqrt7}=\dfrac{\sqrt{3}}{2}=0.866~approx.\)
and does not equal 12/16.

Answer 3

Thank you so much! I totally understand it now @mathmate

Answer 4

Are the answers one of the choices?

Answer 5

Yes it was one of the choices! @mathmate

Answer 6

The part y/x=12/16 is superfluous and incorrect. Was it part of the question?

Answer 7

No it wasn't I was trying to set up a proportion. That is why I couldn't get the answer. @mathmate

Answer 8

ok, that's better.
Have you done metric relations before?

Answer 9

No I have not. @mathmate

Answer 10

Do you have a few minutes for me to show a few more relations related to the question?

Answer 11

Yeah I do! That would be great! @mathmate

Answer 12

I'll start with a diagram.

Answer 13

Metric relations work when (and only when) we drop a perpendicular from a right triangle, like in the diagram of your question.

Answer 14

Now we name the sides.

Answer 15

So far so good?

Answer 16

I am going to take notes actually! Yes so far so good

Answer 17

There are three relations.
1. \(x^2=a(a+b)\) and similarly \(y^2=b(a+b)\)
(we have used these to solve your problem.

Answer 18

Yes i noticed that

Answer 19

2. \(h^2=ab\)
This can come in handy.

Answer 20

3. \(xy=h(a+b)\)
Can you figure out why this works?

Answer 21

...i.e. what does #3 mean geometrically?

Answer 22

because you are just exchanging the values?

Answer 23

It means that x times y = the height times the bottom length ?

Answer 24

#3 works because we are using two different ways to calculate twice the area of the triangle, ab/2=area of triangle, and (a+b)h/2 is just half base * height.

Answer 25

Okay I understand

Answer 26

These are the three metric relations, and they can come in useful in your math toolkit!
Good luck!

Answer 27

Thank you so much. You are so helpful. If you don't mind I will be asking you for help in the future. I gave you a great review. Thanks again! @mathmate

Answer 28

Thank you, and you're welcome any time! :)