<p>Both from test 1</p>
<p>section 3, number 19:
The pyramid shown above has altitude h and a square base of side m. The four edges that meet at V, the vertex, each have length e. If e = m, what is the value of h in terms of m?</p>
<p>The answer is a) m/(sqrt(2))</p>
<p>but I got b) m sqrt(3) / 2</p>
<p>Don't you just do pythagorean theorem?
(.5m)^2 + h^2 = e^2?
I'm confused :o</p>
<p>Then...
Section 7, number 17!</p>
<p>Any help is appreciated :)
Thanks guys!</p>
<p>For the first problem, look carefully at the right triangle to which you’re applying the Pythagoras theorem. The hypotenuse is an edge (length m), and the base is 1/2 the diagonal of the square. You have to apply the Pythagoras theorem twice. First to obtain the base of the triangle (which is 1/2 the diagonal of the square), and a second time to obtain the height.</p>
<p>The diagonal of the square is m sqrt(2). The base is the 1/2 of that. Now apply the Pythagoars theorem to get the height of the traingle. You get height = sqrt (m<em>m - m</em>m/2).</p>
<p>Your error is that you took 1/2 the side of the square as the base of the triangle instead of the 1/2 the diagonal.</p>
<p>Ohh, thanks fogcity!
I did think that at first when doing the problem. Then I think my laziness kicked in and I convinced myself that CB wouldn’t throw something like that at me hehe.</p>
<p>Any takers on the second?</p>
<p>Still need help on the second problem please :)</p>
<p>The two overlapping triangles are triangles ADF and BCE. Because DE=EF=10, the side length of triangle ADF is 10+10=20. Let the point of intersection of lines BE and AD be G. Triange GDE is also and equilateral triangle because it was formed by two overlapping equilateral triangles and it has a side length of 10 (DE=10). Because AD=20, we want AG, which is half of AD or 10. Similarly, BG=10. Thus, our answer is CD+DE+EF+FA+AG+GB+BC=10+10+10+20+10+10+20= 90</p>