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23 Apr 2019, 01:42
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C
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70% (02:29) correct
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In triangle ABC, AB = AC = 28 and BC = 20. Points D, E, and F are on sides AB, BC, and AC, respectively, such that DE and EF are parallel to AC and AB, respectively. What is the perimeter of parallelogram ADEF ?
(A) 48
(B) 52
(C) 56
(D) 60
(E) 72
Attachment:
ec45f0635e288a5e5da73a19319f4eb7045b5eab.png [ 15.31 KiB | Viewed 13307 times ]
firas92
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neha283
By similarity, ABC, BDE & CEF are all isosceles triangles
So \(AB=AC\), \(BD=DE\) & \(EF=CF\) and from parallelogram, \(AD=EF\), \(DE=AF(=BD)\)
Perimeter of parallelogram ADEF = \(AD+DE+EF+AF = AD+BD+AD+BD\)
We know that \(AD+BD = AB = 28\)
So \(Permieter = AB+AB = 28+28= 56\)
(C)
Kudos if this helped !
General Discussion
23 Apr 2019, 23:30
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First imagine points D, E, and F are midpoints of sides AB, BC, and AC. According to the triangle midsegment theorem, the length of the midsegment of a triangle is always half of the length of the side to which it is parallel. If that's the case DE and EF are both 14 and since opposite sides of a parallelogram are also equal that means AF and AD are also 14. That means the perimeter of parallelogram ADEF is 56. But in this question it does not say D, E and F are midpoints of the sides AB, BC, and AC. So how do we figure out the length of DE and EF? Answer is we don't need the individual length of DE and EF. Since AB = AC, if we want to make DE and EF parallel to AC and AB we have to increase and decrease the length of DE and EF in a way that the summation of their length adds up to 28. For example Let's say DE is 3/4th of the length of AC which means DE = 21. If that's the case then EF must be 1/4th of the length of AB which is 7. So perimeter is still 2(21+7) = 56. Hope this helps.
