Thread: Isosceles right triangle problem

Isosceles right triangle problem (solids with know cross-sections)

I have a isosceles right triangle with a known base (square root of (4-x^2) how can I find the height? Do I need to use a law of sin/cos or is there an easier way?

The height is the length of one of the equal sides the area is 1/2s^2

since it is a rt triangle s^2 +s^2 = h^2

s= h/sqrt(2)

in your eg h = 4-x^2

Hello, icu812!

Did you make a sketch?

I have a isosceles right triangle with a known base:\ .

How can I find the height?

Code:

                  *
                * | *
              *   |   *
            *     |b/2  *
          *       |       *
        * 45°     |     45° *
      *  *  *  *  *  *  *  *  *
      : - b/2 - - : - - b/2 - :
      : - - - - - b - - - - - :

Got it?

Yes thanks a lot. It has been probaly 15 years since I had geometry. We just did solids of revolution, and started solids with known cross-section today.