a triangle's area
if you got a triangle which sides squares are 74, 116 and 370, can you figure out the area of the triangle?
posted by Carc at
2:13 PM
Thursday, February 11, 2010
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10 Comments:
This post has been removed by the author.
A=0.25*sqrt((a^2+b^2+c^2)^2-2*(a^4+b^4+c^4))
A=0.25*sqrt((74+116+370)^2-2*(74^2+116^2+370^2))
A=0.25*sqrt(560^2-2*(155832))
A=0.25*sqrt(313600-311664)
A=0.25*sqrt(1936)
A=0.25*44
A=11
Answer
A=11
Cam
This post has been removed by the author.
Alternate
Let 116 be base
orient it parallell to x-axis
c^2=a^2+b^2-2*a*b*cos C
370=74+116-2*sqrt(74)*sqrt(116)*cos C
180/(-2*sqrt(74*116))=cos C
C= cos inv (180/(-2*sqrt(74*116)))
C=166.2637 deg
180-166.2637 =13.73627 deg
this is the angle between 74 side and x axis
sqrt(74)*sin(13.73627)=2.04265
A=1/2*b*h=0.5*sqrt(116)*2.04265
A=0.5*10.77033*2.04265
Answer:
A=11
Cam
Ragknot
We are given the squares of the sides, not the side lengths
i.e. a^2=74, b^2=116, c^2=370
the side lengths are:
a=sqrt(74)
b=sqrt(116)
c=sqrt(370)
Cam
This post has been removed by the author.
Thanks Cam for explaining it. I copied the numbers down and drew the diagram, hurrying to be first.
Well I was first, but wrong.
Congrats for being the first right.
Hi Cam. Thanks for that. I didn't know that equivalent of Heron's formula.
Hi, I get 71.8
sorry boris, the right answer is 11
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