What is the area of triangle FGH?

[AAPL]

In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If DE=5 and GH=15. What is the area of the triangle FGH?

(A) 240
(B) 270
(C) 300
(D) 330
(E) 360

The OA is B.

Experts, I need your help with this PS question, I have many difficults to solve it. I don't know what can I do to solve, I'm really confused!

Is the small triangle equivalent to the large triangle? I don't understand this PS question.

ED it's the side on the right of the figure and GH is on left. Sorry but I don't know how can I do to the image is shown more large!

[GMATGuruNY]

In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If DE=5 and GH=15. What is the area of the triangle FGH?

(A) 240
(B) 270
(C) 300
(D) 330
(E) 360

The following steps refer to figure 1 below:
An inscribed angle that intercepts the diameter of a circle is has a degree measurement of 90 degrees.
Thus, inscribed ∠DEF is a right angle, implying that x=90.

Since line EG is a traversal through parallel lines GH and ED, alternate interior angles GHF and DEF must be equal.
Thus, ∠GHF = ∠DEF = x.






The following steps refer to figure 2 below:
Since ∠GFH and ∠EFD are vertical angles, they must be equal.
Let ∠GFH = ∠EFD = z.
Since ∆FGH and ∆DEF have two pairs of equal angles -- x and y -- the remaining two angles in ∆FGH and ∆DEF must also be equal.
Let ∠HGF= ∠EDF= y.
Since ∆FGH and ∆DEF have the same combination of angles -- x, y and z -- they are SIMILAR.






The following steps refer to figure 3 below:
Since DE=5 and DF=13, ∆DEF is a 5-12-13 triangle.
Thus, EF=12.

Corresponding sides in similar triangles are always in the SAME RATIO.
Since GH and DE are both opposite an angle with a degree measurement of z, they are CORRESPONDING SIDES.
Since GH=15 and DE=5, GH = 3(DE).
Implication:
Every side in ∆FGH must be 3 times the corresponding side in ∆DEF.

Since HF and FE are both opposite an angle with a degree measurement of y, they also are CORRESPONDING SIDES.
Since every side in ∆FGH must be 3 times the corresponding side in ∆DEF, and FE = 12, we get;
HF = 3(FE) = 3(12) = 36.
Thus:
Area of ∆FGH = (1/2)bh = (1/2)(FH)(GH) = (1/2)(36)(15) = 270.


The correct answer is B.