Finding area of a triangle using only Pythagorean theorem
- Thread starter sel147
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Yes; sqrt(161) = ~12.7
Now look at your diagram: do you not see that 12.7 is IMPOSSIBLE?
Where did you get this problem?
Are you sure you copied it properly?
If you want more discussion on your diagram,
please label the 3 points where the 90degree angles are,
so we can refer to them: use D on AB, E on BC extended, F on AC extended.
Getting a headache...bye...
Steven G
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Oh my goodness, Denis with a headache. You normally are a headache on this forum. I can only imagine what will happen tonight.
Otis
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It works for me.
sqrt(161)^2 + 8^2 = 15^2
Are you confusing a leg for the hypotenuse, maybe?
Agree! (I don't think the OP created that diagram.)
I'd already labeled my copy, using your conventions.
I also labeled unknown sides:a = AC
b = AD
d = EC
Side DB is 10∙sqrt(2)
I wrote a system of four equations, using the Pythagorean Theorem for each. I started to solve it by hand, but it became radically unwieldy. (I asked software to solve it, instead.)
Edit: Dang! I made a cut-n-paste error. After fixing that, the software finds no solution.
PS: We can't assume that a diagram is drawn to scale, unless we're told. I've learned this the hard way -- especially with some tricky triangle problems -- where diagrams are meant only to provide labeled info, not true representations. Taken at face value, a given diagram's shapes can be misleading. (At first, I caught myself assuming angle ACB was 90°.)We must be careful. :cool:
As far as other countries using dots to indicate right angles instead of angle congruency, well, that's one more thing for us to remember!
Last edited:
pka
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I think that the confusion is much deeper than that.
NO!! sqrt(161) is the correct calculation;
what I'm trying to f'n say is "it does NOT fit in that diagram"!!
Steven G
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Why does it not fit in the diagram? What EXACTLY does that mean? Do you mean the scale will be off? What the hay do you mean?
Since CD=5 and BC=15 are givens,
isn't c = sqrt(15^2 - 5^2) = ~14.14
The one you're showing evaluates to ~15.16
I'm certainly missing something...
Otis
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Frickin' typos! I botched my data entry.
Yes, side DB (c) is 10∙√2
After fixing my goof, the software reports no solutions. Here's the system:
a^2 = b^2 + 5^2
a^2 = d^2 + 6^2
(b + 10∙√2)^2 = (a + √161)^2 + 8^2
(b + 10∙√2)^2 = (d + 15)^2 + 6^2
I'll take another look, tomorrow …
This is what I've been saying all along: NO SOLUTION!Frickin' typos! I botched my data entry.
Yes, side DB (c) is 10∙√2
After fixing my goof, the software reports no solutions. Here's the system:
a^2 = b^2 + 5^2 [1]
a^2 = d^2 + 6^2 [2]
(b + 10∙√2)^2 = (a + √161)^2 + 8^2 [3]
(b + 10∙√2)^2 = (d + 15)^2 + 6^2 [4]
Jomo and pka: you can now get off my back!!
Otis, you have 4 equations, 3 unknowns.
Solve using [1],[2],[3]: a = ~22.638, b = ~22.079, d = ~21.828
Solve using [1],[2],[4]: a = ~6.315, b = ~3.858, d = ~1.971
So, of course, no solution using all 4 at once!
The real problem is if BF = 8, then impossible for AE = the given 6:
didn't calculate it exactly yet, however AE = ~14
Drawn to scale (as much as possible):
AB = ~43, AD = ~28, AE = ~14, CE = ~25, angleBAE = ~70
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One possible piece of advice for the OP. It is most certainly part of the learning of mathematics to learn to trust your own work. In this case, no matter how convincing the drawing, it may actually fail to exist. As you work through the details, and you discover that things are not working out, you have a choice.
1) I must have done something wrong. Start over or ask for help.
2) I can trace my work. I have performed admirably. The model is inconsistent and must be rejected as impossible.
Guess how many early students don't even know they can pick #2?
Ahhhh gezzzzzzus, I feel like vomiting...
Just realised that the 2 givens 6cm and 8cm are not part
of a f'n diagram, but simply the heights of that triangle
from 2 of the bases. D*mn, all that wasted time
So the area is simply 45.
I ain't looking at another triangle problem for a good while!
Just realised that the 2 givens 6cm and 8cm are not part
of a f'n diagram, but simply the heights of that triangle
from 2 of the bases. D*mn, all that wasted time
So the area is simply 45.
I ain't looking at another triangle problem for a good while!
Ahhhh gezzzzzzus, I feel like vomiting...
Just realised that the 2 givens 6cm and 8cm are not part
of a f'n diagram, but simply the heights of that triangle
from 2 of the bases. D*mn, all that wasted time
So the area is simply 45.
I ain't looking at another triangle problem for a good while!
Thanks a lot. Thank you everyone for your time and effort.If I ever meet you, I'll hit you on the head with my hockey stick!
If I ever meet you, I'll hit you on the head with my hockey stick!
I'd be thankful if you did