If the sum of the lengths of the hypotenuse and another side of a right angled triangle is given then the area of the triangle is maximum when the angle between those sides is
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the answer is 45 degrees.
tiwariaridhima:
Thanx for answer
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let hypotenuse = h
base=x
so, h+x=a
a=constant
laet area be A then,
A=1/2(x*(√h*h-x*x)
squaring both side
A*A={ (a*a)(x*x)-2a(x*x*x) }/4
then differentiate
2dA/dX=1/4{2a*a*x-6*a*x*x}
for maxima
dA/dX=0
then 1/8A(2*a*a*x-6*a*x*x)=0
x=a/3
then cosΘ=x/a-x
cosΘ=(a/3)/(a-a/3)
cosΘ=1/2=cosπ/3
Θ=π/3
base=x
so, h+x=a
a=constant
laet area be A then,
A=1/2(x*(√h*h-x*x)
squaring both side
A*A={ (a*a)(x*x)-2a(x*x*x) }/4
then differentiate
2dA/dX=1/4{2a*a*x-6*a*x*x}
for maxima
dA/dX=0
then 1/8A(2*a*a*x-6*a*x*x)=0
x=a/3
then cosΘ=x/a-x
cosΘ=(a/3)/(a-a/3)
cosΘ=1/2=cosπ/3
Θ=π/3
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