Question 820311
It will help if you draw a diagram:<ol><li>Draw an isosceles triangle and label the vertices so that AB = AC.</li><li>Label side AB as 3s and side AC as 3s. Label BC as s.</li><li>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).</li><li>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.</li></ol>Using the Pythagorean Theorem on one of the right triangles we get:
{{{(s/2)^2+h^2=(3s)^2}}}
Simplifying:
{{{s^2/4+h^2=9s^2}}}
Multiply both sides by 4 (to eliminate the fraction):
{{{s^2+4h^2=36s^2}}}
Subtracting {{{s^2}}}:
{{{4h^2=35s^2}}}
Dividing by 4:
{{{h^2=(35s^2)/4}}}
Square root of each side:
{{{sqrt(h^2)=sqrt((35s^2)/4)}}}
(We will not use the <u>+</u> because h is a height which should never be negative.) Simplifying...
{{{h=(sqrt(35)sqrt(s^2))/sqrt(4))}}}
{{{h=(sqrt(35)*s)/2}}}
which can be rewritten as:
{{{h=(sqrt(35)/2)s}}}<br>
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:
{{{(1/2)s*(sqrt(35)/2)s}}}
which simplifies to:
{{{(sqrt(35)/4)s^2}}}<br>
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.