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# In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that

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Joined: 02 Sep 2009
Posts: 59589
In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that  [#permalink]

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20 Mar 2019, 23:11
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Difficulty:

65% (hard)

Question Stats:

50% (03:05) correct 50% (03:01) wrong based on 24 sessions

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In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of triangle AEB.

(A) 10
(B) 21/2
(C) 12
(D) 25/2
(E) 15

Attachment:

8b4fceed0d7e03e19b1507f86778bc77a8b7f334.png [ 8.38 KiB | Viewed 585 times ]

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Re: In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that  [#permalink]

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20 Mar 2019, 23:44

Solution

Given:
• In rectangle ABCD, AB = 5 and BC = 3.
• Points F and G are on CD so that DF = 1 and GC = 2.
• Lines AF and BG intersect at E.

To find:
• The area of triangle AEB.

Approach and Working:
The length of FG = 5 – (1 + 2) = 2

As FG||AB, we can say ▲EFG and ▲EAB are similar to each other.

If we assume the height of ▲EAB is h, then height of ▲EFG is h – 3.
Therefore, from the similarity, we can say
• $$\frac{FG}{AB} = \frac{h – 3}{h}$$
Or, $$\frac{2}{5} = \frac{h – 3}{h}$$
Or, 2h = 5h – 15
Or, 3h = 15
Or, h = 5

Area of ▲AEB = ½ * 5 * 5 = 25/2

Hence, the correct answer is option D.

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Re: In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that  [#permalink]

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21 Mar 2019, 00:43
find area of trapezium FGAB = 1/2 * ( 2+5) * 3 = 21/2
now using similarlty property of triangle determine the height of the
∆EAB ~ ∆EFG
so
FG/AB = EG/ED
2/3 = H-3/H
er get H = 5
so area of ∆EFG
1/2 * 2*2 = 2
so total are of 21/2 + 2 = 25/2
IMO D

Bunuel wrote:

In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of triangle AEB.

(A) 10
(B) 21/2
(C) 12
(D) 25/2
(E) 15

Attachment:
8b4fceed0d7e03e19b1507f86778bc77a8b7f334.png
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Re: In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that  [#permalink]

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21 Mar 2019, 22:28
Triangles AEB and FEG are similar triangles
Let triangle FEG height be x , then height of AEB = x+3

==> FG/AB = x/x+3

==> 2/5 = x/x+3
==> x = 2

Height of triangle AEB = 5

==> Area of triangle AEB = 5*5 / 2 = 25/2 Option D
Re: In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that   [#permalink] 21 Mar 2019, 22:28
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