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Mathematics 15 Online
OpenStudy (anonymous):

solve

OpenStudy (anonymous):

OpenStudy (anonymous):

how can you know such a thing?

OpenStudy (anonymous):

i didn;t get u @satellite73

OpenStudy (phi):

the best you could do is make the area a function of one side.

OpenStudy (anonymous):

but only hypotenuese is given

OpenStudy (mertsj):

I think they might want an answer in terms of the legs so in this right triangle: |dw:1365104034385:dw|

OpenStudy (mertsj):

We know that a^2+b^2=82^2 so \[a=\sqrt{82^2-b^2}` and` b = \sqrt{82^2-a^a}\]

OpenStudy (mertsj):

And of course a and b are the base and the height so Area = \[A=\frac{\sqrt{(6724-a^2)(6724-b^2}}{2}\]

OpenStudy (anonymous):

i worked it out with my cousin and we got 810

OpenStudy (anonymous):

yes ur right @123456man but how?

OpenStudy (anonymous):

you only know the hypotenuse, you need the base and the height, and you do not have enough information

OpenStudy (anonymous):

if the base and height are the same, then they are both \(41\sqrt{2}\) and so you can compute the area and get 1681

OpenStudy (anonymous):

my cousin did most of the work after we argued about who was right.

OpenStudy (anonymous):

but they don't have to be the same

OpenStudy (anonymous):

the base could be \(1\) and the height could be \(\sqrt{81}\) for example

OpenStudy (anonymous):

he flipped the triangle and used the cosine rule since the hypotenuse is known

OpenStudy (anonymous):

then the area would only be \(\frac{\sqrt{81}}{2}\)

OpenStudy (anonymous):

at the risk of repeating myself, there are infinitely many triangle with hypotenuse 82

OpenStudy (mertsj):

What cosine rule did he use?

OpenStudy (anonymous):

so his answer is wrong?

OpenStudy (mertsj):

Not according to the asker.

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