If [tex]A[/tex] be the area of a right angled triangle and [tex]x[/tex] is one of the sides containing right angle. Prove that the length of altitude on the hypotenuse is
[tex]\dfrac{2Ax}{\sqrt{x^2+4A^2}}[/tex].
[tex]\dfrac{2Ax}{\sqrt{x^2+4A^2}}[/tex].
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Answer:
Step-by-step explanation:
Let say base & height = b & a
Area = (1/2)×b × a
Let say b = x
Then a = 2A/x
Hypotenuse^2 = base^2 + height^2
Hypotenuse^2 = x^2 + (2A/x)^2
Hypotenuse =root (x^2 + (2A/x)^2)
Hypotenuse =(1/x)root (x^4 + 4A^2)
Let say length of Altitude on hypotenuse = y
Then area of triangle = (1/2) × hypotenuse × altitude
A = (1/2) × hypotenuse × y
So y = 2A/(hypotenuse
y = 2A / (1/x)(root (x^4 + 4A^2 )
y = 2Ax/(root(x^4 + 4A^2))
In question its mentioned x^2 instead of x^4 .
Probably typing mistake