<p>Just had a two more questions regarding Math from the BB and was wondering if some one could help me out...would be much appreciated :). They're both from Practice Test ONe. </p>
<p>{Pictures on page 400}</p>
<ol>
<li><p>In right triangle ABC shown above line EF is parallel to line AC and F is the midpoint of BC. What is the area of the shaded rectangular region? *The picture is of two squares in a large triangle with sides of (10)(square roots of 2)
a) 25
b) (25)(square root of 2)
c) 50
d) 50(square root of 2)
e) 100</p></li>
<li><p>The picture 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
a) m/(square root of 2)
b) (m*square root of 3)/2
c) m
d) (2m)/(square root of 3)
e) (m)(square root of 2)</p></li>
</ol>
<p>For 17, all of the triangles you see (the big ones, the small ones, the big collective one, etc.) are all 45-45-90. So, since you know FC = 5sqrt(2) (since F is the midpoint of BC), you know that the side that goes from F to AC of the shaded rectangle = 5 (since that side is the leg of a 45-45-90 triangle with hypotenuse 5 root 2). Now, find EF: Since BF = 5 sqrt(2), and Triangle BEF is 45-45-90, you know that EF = 5 sqrt(2) * sqrt(2), or simply 10. Now, multiple those two sides (5 and 10) together to find the area, 50.</p>
<p>For 19, simply replace e with “m” and draw a triangle from V to the center of the square base (call that center point “C”) to the nearest base vertex to the right (call that base vertex “A”). Now we have triangle VCA, with side VA = m and side VC = h. We want to find h in terms of m, so we need to find side AC. Since AC is 1/2 of the diagonal of the base square, and since the diagonal of a square is simply the hypotenuse of a 45-45-90 triangle, we know that the diagonal is m<em>sqrt(2) (that is, the length of a side of the square times the square root of 2). Since C is the center of the square, we know that AC is 1/2 the diagonal, or m</em>sqrt(2)/2. Now, it’s simple algebra: (m^2)-(m*sqrt(2)/2)^2 = h^2
m^2 - 2(m^2)/4 = h^2
m^2 - (1/2)(m^2) = h^2
(1/2)m^2 = h^2
sqrt((1/2)(m^2)) = sqrt(h^2)
<a href=“m%5E2”>b</a>/sqrt(2) = h**</p>
<p>Sorry if I didn’t explain that first one so well, it’s hard to do without being able to point to the diagram!</p>
Thanks!</p>