SOLUTION: A right triangle has one vertex on a graph of y=√X at (x,y) another at the origin, and the third on the (positive) y-axis at (0,y). Express the area A of the triangle as a fu

Algebra ->  Sequences-and-series -> SOLUTION: A right triangle has one vertex on a graph of y=√X at (x,y) another at the origin, and the third on the (positive) y-axis at (0,y). Express the area A of the triangle as a fu      Log On


   

Question 1126429: A right triangle has one vertex on a graph of y=√X at (x,y) another at the origin, and the third on the (positive) y-axis at (0,y). Express the area A of the triangle as a function of x. Sketch a picture
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The right angle is at (0,y) = (0,sqrt(x)).

One leg is from (0,sqrt(x) to (0,0); length = sqrt(x).
The other leg is from (0,sqrt(x) to (x,y) = (x,sqrt(x)); length x.

The area of the triangle is half the product of the lengths of the legs:

A+=+%28x%2Asqrt%28x%29%29%2F2

You can draw the sketch.