<p>I'm having trouble with number 2 and 10 in the geometry section. It's pretty strange cause i never had trouble with geometry before lol. The answers are at the bottom of the pdf but could someone explain how to do those specific numbers please?</p>
<p>Thanks</p>
<ol>
<li><p>Law of cosines to find length AC. Then draw a line from B to AC such that this line is perpendicular to AC. Find this length with trig and you have a base and height for the triangle.</p></li>
<li><p>Hypotenuse = 10, short leg = sqrt(4+a^2), long leg = sqrt(64+a^2)</p></li>
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<p>So, 100 = 4 + 64 + 2a^2
a = 4</p>
<p>^^Another way to do it is to use slope formula.</p>
<p>(y1-y2)/(x1-x2) * (x1-x2)/(y1-y2) This also gives 4, in a weird way, though, so not sure if it’s right.</p>
<ol>
<li><p>Could just drop altitude to side BC. Altitude will be 3 root 3. Area follows directly.</p></li>
<li><p>Drop altitude from (2,a) to hypotenuse. In a right triangle, that altitude is the geometric mean of the two segments of hypotenuse (lengths 2 and 8). Geometric mean = sqrt(2*8) = 4. Length of the altitude is also the Y coordinate of the point in question.</p></li>
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<ol>
<li> You might recognize that the right triangle could be inscribed in a semicircle. The equation of the whole circle is (x-5)^2 + y^2 = 25. With x known, y follows directly.</li>
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<p>Number 2 would never appear on an SAT - it might show up on a subject test. In any case the area of a triangle is one half the product of two sides times the sin of the included angle. In this case A = 1/2 (6)(8)sin 60 = 24sqrt(3)/2 = 12sqrt(3), choice (B).</p>
<p>For number 10 I would just ue the fact that the slopes of perpendicular lines are negative reciprocals of each other. So -2/a = -a/8. Cross multiplying gives a^2=16. So a=4, choice (B).</p>
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<p>I think it could appear on the SAT. Or, at least on the pre-1994 SAT. :)</p>
<p>It is hard, but doesn’t require trig; like $0.02 said, just draw the altitude from A to side BC, and you have a 30-60-90 triangle.</p>
<p>@fig</p>
<p>You’re right. I spoke too soon. There is a small possibility that something like this could show up as a Level 5 problem on an SAT.</p>