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OpenStudy (anonymous):
Given:
Area of the triangle is 40 and
its sides are 40,35,55 cm.
find the length of the longest altitude.
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OpenStudy (gorv):
1/2*base*height=40
OpenStudy (gorv):
calculate height
OpenStudy (anonymous):
base=?
OpenStudy (anonymous):
\[A=\frac{1}{2}ab\sin\theta\]
OpenStudy (anonymous):
\[\cos\theta=\frac{a^2+b^2-c^2}{2(a)(b)}\]
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OpenStudy (gorv):
put all the length in place of base
the maximum value of height willbe your answer
OpenStudy (anonymous):
@gorv A=1/2 bh only works for right angled triangles. >.>
OpenStudy (gorv):
@Azteck
we have to calculate height and it works for mst of the triangls
okkk if i m wrong tell altitude by ur formula??
OpenStudy (anonymous):
But it also applies if you know which side is the base.
OpenStudy (anonymous):
@Azteck A=1/2 bh works for any type of triangle......only if you are given the required lengths......
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OpenStudy (gorv):
thats wat i m saying
the side which will give max altitude will be considered as base
OpenStudy (gorv):
35 will give max height so take it as base
OpenStudy (anonymous):
how 35 cmes base
OpenStudy (gorv):
put all the sides in formula told
base =35 will give max altitude that is height
so side=35 will be consideredas the base
OpenStudy (anonymous):
k
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OpenStudy (anonymous):
i got it
OpenStudy (anonymous):
@gorv thanks bro
OpenStudy (gorv):
take same approch
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