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In the figure, AC = 3, CE = x, and BC is parallel to DE. If the area of triangle ABC is 1/12 of the area of triangle ADE, then x = ?
A. \(3\sqrt{3}-3\)
B. \(3\)
C. \(6\)
D. \(6\sqrt{3}-3\)
E. \(6\sqrt{3}\)
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04 Feb 2012, 18:09
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Another useful property: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).
Next, note that ABC and ADE are similar triangles, thus the ratio of their areas is the square of the ratio of their sides:
- \(\frac{AE^2}{AC^2}=\frac{12}{1}\);
\(\frac{AE}{AC}=\sqrt{12}=2\sqrt{3}\);
\(AE=2\sqrt{3}*AC=6\sqrt{3}\);
\(x=AE-AC=6\sqrt{3}-3\).
For more on triangles check Triangles chapter of Math Book: https://gmatclub.com/forum/math-triangles-87197.html
Hope it helps.
04 Feb 2012, 18:19
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enigma123
Similar triangles are triangles in which the three angles are identical. Trianlges ABC and ADE share one common angle D and also since BC is parallel to DE, <ABC=<ADE and <ACB=<AED.
Notice that it is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
Another useful property of similar triangles: in similar triangles, the sides of the triangles are in some proportion to one another. For example, a triangle with lengths 3, 4, and 5 has the same angle measures as a triangle with lengths 6, 8, and 10. The two triangles are similar, and all of the sides of the larger triangle are twice the size of the corresponding legs on the smaller triangle.
Hope it helps.
