<p>Blue book 2nd edition: 1st test, 3rd section,page 401, number 19</p>
<p>h^2+ (m^2/2)^2=m^2</p>
<p>how do that solve h in term of x?</p>
<p>you’ve got the problem set up wrong.
there are two seperate triangles in the problem. Lets say the unknown hypotenuse of the triangle that has h in it is A. Since this value is unknown for both triangles, we have to solve for that first before we solve for h.
since e=m, e^2-(m/2)^2=A^2 turns into m^2-(m/2)^2=A^2. solving for A gives you root3(m)/2.
now plug in this value for A into the triangle with h as a value. H^2+(m/2)^2=A^2 turns into H^2+(m/2)^2=(root3(m)/2)^2, which becomes H^2=(3m^2/4)-(m^2/4).
H then equals root2(m)/2, which is the same thing as saying A. m/root2</p>
<p>^ What. The. ****? Dude. Plug-in numbers. Algebra is way too time-consuming.</p>