SOLUTION: 1. what is the area of the isosceles triangle 2 sides each measuring 3s with a base s ? 2. solve = 5/(4a-7bi)

Question 820311: 1.
what is the area of the isosceles triangle 2 sides each measuring 3s with a base s ?
2. solve = 5/(4a-7bi)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
It will help if you draw a diagram:

Draw an isosceles triangle and label the vertices so that AB = AC.
Label side AB as 3s and side AC as 3s. Label BC as s.
Draw a perpendicular from A to BC. Label the point where it intersects with BC as D. And label its length as h (for height).
Because the two right triangles, ADB and ADC, have the same hypotenuse, 3s, and the same leg, AD, they are congruent. This makes BD = CD. Since BC is s and BD = CD, BD = CD = s/2.

Using the Pythagorean Theorem on one of the right triangles we get:

Simplifying:

Multiply both sides by 4 (to eliminate the fraction):

Subtracting :

Dividing by 4:

Square root of each side:

(We will not use the + because h is a height which should never be negative.) Simplifying...

which can be rewritten as:

The area of the isosceles triangle ABC is 1/2 times the base times the height. Using a base of s and the height we just found we get:

which simplifies to:

P.S. Problem 2 is not correct. Expressions are not "solved". Either the problem is incomplete or you did not include the proper instructions for the problem. Plus, it really doesn't belong as a posting in the radicals category. (It belongs under Complex Numbers. Please re-post the full, correct problem under Complex Numbers.
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