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:
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