geometry help!
crapp, i thought it'd be easy stuff.
but i can still help: studied trig in 6th
1. the last one, i think
2. is the second one, i think too
:D
not 100% sure, but pretty sure
like 98%
someone school me plz i need halp
No. 1. Is wrong. Use the pythagoereom theorem. a^2 +b^2 =c^2 Where a and b are the legs of the right triangle And c is the hypotenuse You know one of the leg is 18 and the hypotenuse is 21 So 18^2 + b^2 =21^2 Can you solve for b?
it'd have to be the square root of something
117
but how would you convert that into a square root?
first is A then?
Sqrt of 117 You know 117=9x13 And 9 is a perfect square because 3^2=9 or sqrt9=3 So Sqrt117= 3 sqrt 13
how do i solve the second one?
yeah i will try to get the second one
Have you learned about the similarities in a right triangle? And with that you have The altitude is the geometric mean of the segments into which it separates the hypothenuse
In other words, if we were to use you diagram, it'll be GI= sqrt (FI x IH) Just to let you know, everything on the right side is under the radical sign
Nikato, if you find yourself having trouble explaining, I'd be willing to assist.
would it also be A too? I solved it and 8.4
Yes. It would also be A. But I think I would best it you leave your answers in radical form
oh ok.. thank you for your help!
your welcome!
you did way better than i did... i kind suck at this kinda stuff...
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