Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2006-10-08 21:36:13

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Hard(ish) Geometry Problem

Hi there everyone!

I have this really annoying geometry problem, and it involves similar triangles I think sad. I am still confused.

RQ = 12
PQ = 18
Find the length of UR.

THX

Oh, and can someone please tell me how to type using complex notation? e.g putting in fractions, functions, exponents XD

Last edited by Toast (2006-10-08 22:47:44)

Offline

#2 2006-10-08 23:21:58

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: Hard(ish) Geometry Problem

First, note that:

by pythagoras.

Now, let the angle RPQ be called

.

Then:

simply by the definition of tan and arctan.

Now we see that:

so:

So we can get the length VR by using the lengths for PV and PR:

(remember that we do know what

is, I'm just not using it for neatness right now)

Now note that the angle SRP is the same size as the angle RPQ.

Using this, we can see that:

Since we have VR and we have

, you can calculate UR. I'll leave that to you though.



As for the complicated notation, have a look in this thread.


Bad speling makes me [sic]

Offline

#3 2006-10-08 23:35:22

Devantè
Real Member
Registered: 2006-07-14
Posts: 6,400

Re: Hard(ish) Geometry Problem

This would be the kind of problem that could be put in 'Exercises' because it's a nice and easy one.

Offline

#4 2006-10-09 00:19:41

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: Hard(ish) Geometry Problem

Wow, thanks for the advice - i underestimated the use of that angle and cos, infact I neglected trigonometry, thinking that similar triangles would suffice.

Oh, and Devante, i've got quite a few more of these kinds of problems, if you want to put them on the website big_smile

alright kthxbye

Offline

#5 2006-10-09 00:21:37

Devantè
Real Member
Registered: 2006-07-14
Posts: 6,400

Re: Hard(ish) Geometry Problem

It'll definitely qualify for the Exercises forum. smile

Offline

#6 2006-10-09 04:04:09

pi man
Member
Registered: 2006-07-06
Posts: 251

Re: Hard(ish) Geometry Problem

You can also use similar triangles to solve this as you suggested.    (I assume that your figure is a rectangle)

First, figure out the lenght of PR as Dross pointed out.   Triangle PQR is similar PVQ because two of the angles are equal - both are right triangles and angles  PVQ and RPQ are obviously equal since they are the same angle.    Because they are similar: 

You know the values for PQ (18) and for RQ (12), so you can solve for PV.   Then you can subtract the length of PV from PR to get VR.   

Next, the triangle RVU is similar to PQR.   They are both right angles and angle RPQ  = URV (alternate interior angles).  So:

You know PQ, PR and VR so you can solve for UR.

Offline

Board footer

Powered by FluxBB