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D
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Difficulty:
55%
(hard)
Question Stats:
63% (01:55) correct
37%
(02:01)
wrong
based on 357
sessions
History
Date
Time
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Not Attempted Yet
Attachment:
Untitled.png [ 24.36 KiB | Viewed 15081 times ]
(1) Side DC = 20
(2) Side AC = 8
Hi,
In the explanation of this question, it's telling me that the two angles at C are the same and therefore, the triangles are similar. But i'm not understanding why, how do we know thse two angles are equal?
Please help, thank you very much.
In the explanation of this question, it's telling me that the two angles at C are the same and therefore, the triangles are similar. But i'm not understanding why, how do we know thse two angles are equal?
Please help, thank you very much.
21 Feb 2014, 09:08
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What is the area of triangle ABC ?
Angle ACE is a straight line, hence it's 180°. Now, since (angle BCD)=90°, then (angle ACB) + (angle DCE) = 180° - 90° = 90°.
Next, in triangle ABC, (angle ACB) + (angle ABC) = 90°. Thus we have that:
(angle ACB) + (angle ABC) = 90° = (angle ACB) + (angle DCE) --> (angle ABC) = (angle DCE), which on the other hand implies that (angle ACB) = (angle CDE).
Therefore, all three angles in triangles ABC and CDE are equal, so these triangles are similar.
If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
(1) Side DC = 20 --> \(CE=\sqrt{20^2-16^2}=12\) --> \(AC=20-12=8\) --> \(\frac{AREA_{CDE}}{area_{ABC}}=\frac{16^2}{8^2}\). Since the area of triangle CDE = 16*12/2 = 96, then \(\frac{96}{area_{ABC}}=\frac{16^2}{8^2}\) --> we can find the area od triangle ABC. Sufficient.
(2) Side AC = 8. The same info as above. Sufficient.
Answer: D.
Below image might help to understand better:
Attachment:
Untitled.png [ 24.88 KiB | Viewed 13213 times ]
General Discussion
21 Feb 2014, 09:35
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Thank you Bunuel ! Superb explanation.
