<p>As of now I only have one problem, but I'll probably post a few questions later.</p>
<p>Pg. 401 BB2 #19 (It has a diagram but I don't think you will need it)</p>
<p>The pyramid shown above has altitude h and a square base of side m. The four edges that meet at V, the vertex of the pyramid, each have length e. If e = m, what is the value of h in terms of m?</p>
<p>A) m/sqrt 2</p>
<p>B) msqrt 3 /2</p>
<p>C) m</p>
<p>D) 2m/sqrt 3</p>
<p>E) msqrt 2</p>
<p>The answer is A, but as you can tell, I don't know how it was found.</p>
<p>First start out by finding the diagonal of the square base. Since all of the sides equal m, the diagonal = m(sqrt2)</p>
<p>Divide that by two to get m(sqrt2)/2. You have to use this length to create a right triangle composed of m(sqrt2)/2, e, and h. e is the diagonal of this triangle, so by the pythagorean theorem we get h^2 + 2[m^2]/4 = e^2 = m^2 (because e = m) </p>
<p>Simplifying we get h^2 = m^2/2 and finally h = m/sqrt2</p>
<p>nevermind i got it ill help you out tho lockdown</p>
<p>the reason you divide by two is because the diagnol of the base of the pyramid = m sqroot 2 but since you have to draw a right triangle you’re only dealing with half the diagnol </p>
<p>therefore it is ( m sqroot 2 ) / 2</p>
<p>if you didn’t know, log on collegeboard i jus figured this out they have answer explanations for the new book, you just have to go on page 18 and type the code-word or whatever after you log in and select the service.</p>