# Finding Missing Triangle Lengths...

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• Mar 30th 2008, 11:32 AM
zachcumer
Finding Missing Triangle Lengths...
See attached..all I know is that each tick mark represents 6...

Thanks all the help I could get would be great!!!
• Mar 30th 2008, 11:40 AM
well the line passing from one point through the triangle to the base of another side can be calculated using pythagoras theorem

12^2 - 6^2 = (x+y)^2
x+y = sqrt(108)

the gravity point (where all the internal lines meet) is always 2/3 from the apex or 1/3 from the base

therefore

y = (2/3)sqrt(108)
x = (1/3)sqrt(108)
• Mar 30th 2008, 11:47 AM
zachcumer
Thanks for the help!
Thanks for the help...have a nice day!
• Mar 30th 2008, 12:05 PM
zachcumer
More Problems!!
Here is another fun one!

Look at attached...@!@

The tick marks are 9...

Thanks

Best,

Zach
• Mar 30th 2008, 12:06 PM
lol xD same thing just change all the 6 to 9 and all the 12 to 18 (Giggle)
• Mar 30th 2008, 12:09 PM
zachcumer
What I have been getting is: (x + y)^2 + 81 = 18^2...am I right...if so...hmm
• Mar 30th 2008, 12:13 PM
yup thats right xD just continue the other steps

(i made a mistake above (Lipssealed) sorrie it should be 12^2 - 6^2 instead of adding them up)
• Mar 30th 2008, 12:14 PM
zachcumer
Correction!
18^2 - 9^2 = (x +y)^2

x + y = √(243)
• Mar 30th 2008, 12:15 PM
nono o.o"

it is

18^2- 9^2 = (x+y)^2
• Mar 30th 2008, 12:22 PM
zachcumer
got it!

Hey so the final answer is : .....one momento...

x = 1/3√(243)

y = 2/3√(243)

RIGHT?!?

Cool. Thanks..
• Mar 30th 2008, 12:24 PM
xD eres correcto bravo
• Mar 30th 2008, 12:25 PM
zachcumer
Shank You Very Much!!!
• Mar 30th 2008, 12:27 PM
zachcumer
More!! Yaya!!
Check dis out..dawg!

okay..look at attached!
• Mar 30th 2008, 12:34 PM
lol they already given you the length xD

the answer is x = (1/3)(18)
y = (2/3)(18)
• Mar 30th 2008, 12:42 PM
zachcumer
Hey I need to find the area of THIS WHOLE TRIANGLE..
here is what I came up with:

36 + (2/3√(108) + 1/3√(108))^2 = 12^2

?
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