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

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


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
It will help if you draw a diagram:
  1. Draw an isosceles triangle and label the vertices so that AB = AC.
  2. Label side AB as 3s and side AC as 3s. Label BC as s.
  3. 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).
  4. 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:
%28s%2F2%29%5E2%2Bh%5E2=%283s%29%5E2
Simplifying:
s%5E2%2F4%2Bh%5E2=9s%5E2
Multiply both sides by 4 (to eliminate the fraction):
s%5E2%2B4h%5E2=36s%5E2
Subtracting s%5E2:
4h%5E2=35s%5E2
Dividing by 4:
h%5E2=%2835s%5E2%29%2F4
Square root of each side:
sqrt%28h%5E2%29=sqrt%28%2835s%5E2%29%2F4%29
(We will not use the + because h is a height which should never be negative.) Simplifying...
h=%28sqrt%2835%29sqrt%28s%5E2%29%29%2Fsqrt%284%29%29
h=%28sqrt%2835%29%2As%29%2F2
which can be rewritten as:
h=%28sqrt%2835%29%2F2%29s

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:
%281%2F2%29s%2A%28sqrt%2835%29%2F2%29s
which simplifies to:
%28sqrt%2835%29%2F4%29s%5E2

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.