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Offline Munchor

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Re: Geometry Question
« Reply #15 on: May 05, 2011, 04:20:40 pm »
n64_super_mario_64_start.jpg O.O

Very well :)
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Offline Builderboy

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Re: Geometry Question
« Reply #16 on: May 05, 2011, 04:21:55 pm »
Hah whenever I need a fresh image I always find one I'm not using on my desktop and clear it in paint and reuse it :P I don't always remember to rename it too ;D
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Offline Munchor

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Re: Geometry Question
« Reply #17 on: May 05, 2011, 04:22:21 pm »
Quote from: Builderboy on May 05, 2011, 04:21:55 pm
Hah whenever I need a fresh image I always find one I'm not using on my desktop and clear it in paint and reuse it :P I don't always remember to rename it too ;D

I imagine your desktop :P
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Offline Builderboy

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Re: Geometry Question
« Reply #18 on: May 05, 2011, 04:24:57 pm »
Its actually almost empty because I just got this computer O.O
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Offline calcdude84se

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Re: Geometry Question
« Reply #19 on: May 05, 2011, 05:10:51 pm »
Nice, I like the same-area solution :D
Another way is this: (using the labels in Builder's picture)
ΔABC and ΔDBA are similar triangles, so AC/AD=BC/AB.
So AD=AC*AB/BC=40*30/50=24, as the others got. :)
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Offline AngelFish

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Re: Geometry Question
« Reply #20 on: May 05, 2011, 05:32:24 pm »
Quote from: Horrowind on May 05, 2011, 03:49:40 pm
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken

That really only works if the triangle is symmetric across both sides of the "height" line. It's not a general formula for the height of a triangle, especially when the triangle is obtuse (not the case here).
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Offline calcdude84se

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Re: Geometry Question
« Reply #21 on: May 05, 2011, 05:40:07 pm »
It doesn't have to be symmetric, it just has to be a right triangle, AFAICT, which is the fact that my similar-triangles solution relies upon.
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Offline AngelFish

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Re: Geometry Question
« Reply #22 on: May 05, 2011, 05:41:27 pm »
Er, yeah. I forgot that the right angle is a degenerate case that forces the height into discrete values.
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Offline calcdude84se

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Re: Geometry Question
« Reply #23 on: May 05, 2011, 05:51:25 pm »
Oh okay :P We all make mistakes and forget, don't worry. ;D
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Offline Michael_Lee

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Re: Geometry Question
« Reply #24 on: May 05, 2011, 09:18:57 pm »
Quote from: Horrowind on May 05, 2011, 03:49:40 pm
well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leads to h = 24, if i'm not mistaken

And here I was, about jump straight in with law of cosine...
Thumbs up for simplicity!
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Offline calcdude84se

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Re: Geometry Question
« Reply #25 on: May 05, 2011, 09:20:35 pm »
What about me? Or are similar triangles too much? :P
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Offline Michael_Lee

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Re: Geometry Question
« Reply #26 on: May 05, 2011, 09:23:15 pm »
Wait.  How are they similar?

(I'm pretty sure you're going to jump in with an obvious answer that I overlooked...)
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Offline calcdude84se

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Re: Geometry Question
« Reply #27 on: May 05, 2011, 09:24:19 pm »
By AA (angle-angle). They have a right angle and angle B in common.
Edit: angle B, not A. Fixed.
« Last Edit: May 05, 2011, 09:25:29 pm by calcdude84se »
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Offline Michael_Lee

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Re: Geometry Question
« Reply #28 on: May 05, 2011, 09:26:58 pm »
Ohhh...

Okay then, a thumbs-up for you too.
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Offline Stefan Bauwens

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Re: Geometry Question
« Reply #29 on: May 06, 2011, 04:52:17 am »
I finally figured it out today. My mind was to busy yesterday I guess.
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