<p>This is one of the questions from the new test given in BB2. Not sure if a lot of the CC community have the second edition but hopefully some do and can help. </p>
<p>It's one of them figure problems: "The pyramid shown above has altitude ha nd 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>Obviously it's gonna be hard to answer it without the figure but yeah. Maybe I'm going about it the wrong way but I've tried making a triangle from h, vertex, E and a 90 60 30 triangle but can't seem to corrolate m/e to h. </p>
<p>Ok this is hard to explain but ill try my best…</p>
<p>For this kind of a problem you need to realize that to find h is just by simply using the pythagorean theorem since a right triangle forms. The hypotenuse of the triangle is m (it would be e but since e=m, we’ll say that the hypotenuse is m.)
Now here’s where it gets tricky…To find the small leg of the triangle you must first find the hypotenuse of the square base of the pyramid and then divide this hypotenuse by 2 since the leg is only half the hypotenuse of the square base. So the hypotenuse of the leg comes out to be m radical 2 over 2. Then you plug the leg value and the hypotenuse value of the triangle into the pythagorean theorem and solve for h. Which should come out to A.</p>
<p>I don’t get how the hypotenuse can be m radical 2 over 2??? Isn’t the hypotenuse 2m? (square root of m^2 + m^2)???I’m so confused with this lol =(</p>
<p>Oh sorry the hypotenuse is not m radical 2 over 2, the shorter of the two LEGs is m radical 2 over 2. The other leg is h and the hypotenuse is m because it was established that e=m and since e is the edge, which also happens to be the hypotenuse, then e can be substituted for m (since the question asks to find the value of h in terms of m). Then all there’s left to do is plug all this into the pth theorem. (h^2 + (m radical 2 over 2)^2 = m^2))—Solve for h. (answer A)</p>
<p>so half of the hypotenuse of the square base pyramid is square root of (2m^2)/2 right? but how does it turn into m sqr root 2/2??? I don’t understand that lol</p>
<p>Ok…the whole hypotenuse of the square base is m radical 2. since its only half the hypotenuse, its m radical 2 over 2.
What you are doing is also correct but it needs to be simplified more. the squ. root of 2m^2 is m radical 2 because the m squared and the square root cancel each other out leaving m radical 2, which then turns into m radical 2 over 2. Any better?</p>
<p>The triangle you’re looking for is that which has as its legs 1) half a diagonal of the square base and 2) the altitude, h. The hypotenuse is e, which we’ll just call m since e = m. We need to find that half-diagonal.</p>
<p>Draw a square with sides m, and draw two half-diagonals such that a 45-45-90 triangle is created. Then, each smaller side is m/sqrt(2) since the sides of a 45-45-90 triangle are in 1:1:sqrt(2) ratio. Each smaller side is also a half-diagonal, so there you go.</p>
<p>yeah see i was doing all this but the only thign i was doing was trying to use all this info to construct the 30 60 90 triangle which isn’t very helpful when i could’ve and should’ve just done pythagorean… thanks.</p>