A right triangle has one vertex on the graph of y16x2 x0 at

A right triangle has one vertex on the graph of y=16-x^2, x>0 at (x,y) another at the origin, and the third on the positive x-axis at (x,0). Express the area A of the triangle as a function of x.

Solution

---A right triangle has one vertex on the graph of y=16-x^2, x>0 at (x,y)

----another vertex at the origin

---third on the positive x-axis at (x,0)

Base of triangle = x

height = distnac between (0,0) and point (x,y) on curve y = 16 -x^2

height = sqrt(x^2 +y^2) = sqrt( x^2 + (16-x^2)^2 )

So, Area of Triangle = (1/2)*base*height

A = (1/2)(x)sqrt( x^2 + (16-x^2)^2 )

A = [ x*sqrt( x^2 + (16-x^2)^2 )]/2