<p>The figure above shows three squares with sides of length 5, 7, and x, respectively. If A, B, and C lie on line MATHTEXT what is the value of x?
<a href="http://i55.tinypic.com/2ilmv5c.jpg%5B/IMG%5D">http://i55.tinypic.com/2ilmv5c.jpg
</a></p>
<p>The answer is 49/5. How did you get it?
CB solves it by similarity using angle ABD = angle BCE. How are these two angles equal? Can this question be solved using slopes?</p>
<p>“The figure above shows three squares with sides of length 5, 7, and x, respectively. If A, B, and C lie on line L what is the value of x?”</p>
<p>Hey, this is how I would solve it:</p>
<p>When you solve for AB, you get (29)^1/2. Now, since triangles ABD and BCE are similar and 7 is 1.4 times 5, it should follow that distance BC is 1.4 times distance AB. BC turns out to be about 7.54 and with the value of the base, 7, you can solve for the height of the triangle, CE. CE is 2.8. To find the value of x, just add 2.8 to 7. The answer is 9.8 or 49/5. </p>
<p>I hope that helps.</p>
<p>You can use comparison, too. :)</p>
<p>AD/BD = BE/CE
5/(7-5) = 7/(x-7)
5x-35 = 14
5x = 49
x = 49/5</p>
<p>Right, but how do you apply similarity? They both are right angled; what’s the 2nd common angle?</p>
<p>^ They share a common line (Properties of Angles…).</p>
<p>I haven’t done this stuff in a while so I am not a 100% sure but that’s what I used to conclude that the two triangles were similar.</p>
<p>^
Hmm, one angle and one side aren’t sufficient to prove similarity. You need 2 angles, OR, 2 sides and the included angle…</p>
<p>The CB explanation proves similarity by using the Right Angle and angle ABD = angle BCE. I understand the right angle part, but the second part…?
How is angle ABD = angle BCE? I am missing the obvious?</p>
<p>Thanks! :)</p>
<p>I never said that the sides had equal length because they shared a common line. They share an angle. Look at line CE and BD and apply the properties of angles.</p>
<p>parallel lines’ intersection (with the beyond portion chopped off) with a single line creates angles of equal measure</p>
<p>BD and CE are perpendicular to AD and BE respectively, and AD and BE are necessarily parallel (otherwise it’s impossible for BD and CE to both create right angles AND have the 5 and 7 squares share a side). Lines perpendicular to parallel lines are themselves parallel.</p>
<p>then I did what princetondreams did</p>
<p>This is how I did it. Find the hyp of the smallest triangle, which is square root of 29. Then use similar triangles to find the hyp of the second triangle. Then I imagined the two triangles as one big triangle. I added the hyps, along with 7 and 5, then found for side b. After I did that, I added 5 to get 9.8.</p>
<p>Thanks everyone!! But how long did it take you guys to solve this question? It took my like 2 whole minutes. :|</p>
<p>30 seconds…
um
5/7=(7-5)/a => a=2.8
x=7+a=9.8=49/5</p>
