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# In the diagram above, ED is parallel to GH, and the circle has a diame

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In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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08 Mar 2015, 19:27
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In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

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

Kudos for a correct solution.

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In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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08 Mar 2015, 19:40
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Bunuel wrote:
Attachment:
The attachment gpp_img4.png is no longer available
In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

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

Kudos for a correct solution.

Angle FED = 90 degrees since FD is the diameter of the circle. Since GH is || to ED we know that angle GHF is also 90 degrees as well.

We should know at this point that the triangles are similar with all the sides of triangle GHF is three times bigger than the sides of triangle FED.

To figure out the area of triangle GHF we need to figure out the height of the right triangle FED.

FD^2 = ED^2 + FE^2
13^2 = ED^2 + 25
ED = 12

Knowing that FH = 3*FE we can plug in.

FH = 3(12)
FH = 36

Now we can get the area of right triangle GHF.

(B*H)/2 = Area
(15*36)/2 = Area
270 = Area

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Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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08 Mar 2015, 21:17
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ED=5, FD=13, THEN ANGLE AT E 90DEGREE
THEN EF=12

ED PARALLEL TO GH
THEN FOLLOWING SIDES ARE EQUAL IN RATIO
EF/GF=FD/FH=ED/GH

THEN 12/GF=5/15 =>GF=36.

13/FH=1/3=>FH=39.

FH=39, GF=36 AND GH=15

RIGHT ANGLED TRIANGLE. = 1/2*36*15=270
option B sufficient
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Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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08 Mar 2015, 21:28
1
Bunuel wrote:
Attachment:
gpp_img4.png
In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

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

Kudos for a correct solution.

Note that the triangle FED is similar to triangle FHG by AA (angles F of both triangles are vertically opposite angles and EH is a transversal to parallel lines ED and GH so angles E and H are equal because they are alternate interior angles)

Since ratio of their sides is 5:15 i.e. 1:3, ratio of their areas will be 1:9 (square of the ratio of sides)

FD = 13 and ED = 5 in right triangle FED so this is a 5-12-13 triangle. Area = (1/2)*5*12 = 30

Area of triangle FGH = 9*30 = 270

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Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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08 Mar 2015, 23:26
2
Hi

This question test the similar triangles concept

As GF // ED
Therefore Angle marked in the diagram are equal and by AAA Triangles GHF & FED are similar
Now as we know for a pair of similar triangles the ratio of the corresponding sides is equal

Therefore GH/ED = HF/FE
Now HF = 36

Area of triangle = 1/2*15*36 = 270
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Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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09 Mar 2015, 01:35
1

Right triangle DEF is similar to right triangle GHF; Area in the ratio 1:3

Area of right triangle GHF $$= \frac{1}{2} (12*3) * 15 = 270$$
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Posts: 62496
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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15 Mar 2015, 19:30
Bunuel wrote:

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

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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Let’s begin by focusing on triangle FED. The angle ∠E spans a diameter, so ∠E = 90°. Thus, triangle FED is a right triangle with hypotenuse FD = 13 and leg ED = 5. It will save you a tremendous amount of calculations here if you already have memorized the 5-12-13 Pythagorean Triplet. Thus, FE = 12. Area = (0.5)bh = (0.5)(12)(5) = 30.

Because ED and GH are parallel, all the angles are equal, and the two triangles are similar. From ED = 5 to GH = 15 we scale up by a scale factor of k = 3. Lengths are multiplied by the scale factor, and areas are multiplied by the scale factor squared, k^2 = 9. 30*9 = 270 is the area of FGH.

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Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

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09 May 2016, 22:32
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Re: In the diagram above, ED is parallel to GH, and the circle has a diame   [#permalink] 09 May 2016, 22:32
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