Question

If $A$ be the area of a right angled triangle and $x$ is one of the sides containing right angle. Prove that the length of altitude on the hypotenuse is
$\dfrac{2Ax}{\sqrt{x^2+4A^2}}$.​

Answer 1

Let the base be b .

Let the height be h .

A = 1/2 b h

= > b = 2 A / h

Now it is given that x is one of the sides containing right angles.

This means :

Either b = 2 A / x

or h = 2 A / x

Whatever be the case :

Hypotenuse² = height² + base²

any one side is 2 A / x and the other side is x .

Let Hypotenuse be k

k² = x² + ( 2 A / x )²

= > k² = x² + 4 A² / x²

= > k² = ( x⁴ + 4 A² ) / x²

= > k = √( x⁴ + 4 A² ) x

Area of the right angle containing hypotenuse :

A = 1/2 k a

Let altitude on hypotenuse be a

= > 2 A = k a

= > a = 2 A / k

= > a = 2 A / ( √( x⁴ + 4 A² ) x )

= > a = 2 A x / ( √( x⁴ + 4 A² ) )

Hence it is proved .

Answer 2

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