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20 Mar 2019, 23:11
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (03:14) correct
40%
(03:13)
wrong
based on 43
sessions
History
Date
Time
Result
Not Attempted Yet
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 22054 times ]
20 Mar 2019, 23:44
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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.
Answer: D
Archit3110
Major Poster
21 Mar 2019, 00:43
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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
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
