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Mathematics 14 Online
OpenStudy (anonymous):

Geometry pls help!!! An isosceles triangle has area of 105 ft2. If the base is 10 ft, what is the length of each leg? Round your answer to the nearest tenth.

OpenStudy (anonymous):

a.21.6ft b.541ft c.14.6ft d.23.3ft

OpenStudy (solomonzelman):

assuming that the base is not one of the equal sides.

OpenStudy (solomonzelman):

|dw:1387117043671:dw|

OpenStudy (solomonzelman):

\[x^2-5^2=~~hieght\]\[(x^2-25)\times 5~=~52.5\]

OpenStudy (solomonzelman):

\[5x^2-125=52.5\]\[5x^2=177.5\]\[x^2=35.5\]\[x=\sqrt{35.5}\]

OpenStudy (solomonzelman):

See how I did it, I found the height in terms of x.

OpenStudy (solomonzelman):

\[\sqrt{35.5}=\sqrt{335}~\times~1/10\]I think this is it.

OpenStudy (anonymous):

i understand but the answer is none of the given?

OpenStudy (solomonzelman):

Yes, I probably made an error, I'll redo it....

OpenStudy (solomonzelman):

|dw:1387117895716:dw| a.21.6ft b.541ft c.14.6ft d.23.3ft

OpenStudy (solomonzelman):

\[5~\times~h~\times 0.5=~52.5\]

OpenStudy (solomonzelman):

\[h=52.5~\div~0.25~~=~~21\]

OpenStudy (solomonzelman):

\[\sqrt{5^2+21^2}=x\]

OpenStudy (solomonzelman):

\[x=\sqrt{466}≈21.6\]

OpenStudy (anonymous):

thank you!!!

OpenStudy (solomonzelman):

Yes, when I got sqrt{35.5} I was totally fricked out, this time I am sure I am right. Anytime. and ty.

OpenStudy (anonymous):

hey can you pls help me with just one more question?

OpenStudy (solomonzelman):

Yes, but post it in a new thread if you can please.

OpenStudy (anonymous):

ok

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