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17 Oct 2017, 01:48
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C
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Difficulty:
75%
(hard)
Question Stats:
67% (03:35) correct
33%
(03:22)
wrong
based on 76
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CHALLENGE PROBLEM
If EF is parallel to DG, the length of AB is half of that of AC, the length of CG is equal to that of CD, and the length of DE is equal to that of EF, what is the measure of angle DFE in degrees?
(A) 47.5º
(B) 50º
(C) 52.5º
(D) 55º
(E) 57.5º
Attachment:
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Updated on: 18 Oct 2017, 02:06
Originally posted by animesh mohanty on 17 Oct 2017, 04:52.
Last edited by animesh mohanty on 18 Oct 2017, 02:06, edited 1 time in total.
Last edited by animesh mohanty on 18 Oct 2017, 02:06, edited 1 time in total.
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\(\triangle\) ABC is a right angle triangle where AB=(1/2)BC. The angle split would be 30:60:90 in such a case. \(\angle ACB\)=\(30^{\circ}\) .
moving on to \(\triangle\)CDG which is a isosceles triangle \(\angle\) DCG is =\(30^{\circ}\) by using opposite angles property.
Now the rest of two angles \(\angle\) CDG and \(\angle\) CGD will be \(75^{\circ}\).
since EF is parallel to DG so \(\angle\) DEF is also \(75^{\circ}\).
\(\triangle\) DEF is also an isosceles triangle so \(\angle\) DFE= (180-75)/2=52.5
moving on to \(\triangle\)CDG which is a isosceles triangle \(\angle\) DCG is =\(30^{\circ}\) by using opposite angles property.
Now the rest of two angles \(\angle\) CDG and \(\angle\) CGD will be \(75^{\circ}\).
since EF is parallel to DG so \(\angle\) DEF is also \(75^{\circ}\).
\(\triangle\) DEF is also an isosceles triangle so \(\angle\) DFE= (180-75)/2=52.5
17 Oct 2017, 09:31
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Given that AB=1/2AC. Triangle must be a 30-60-90 right angled triangle.
LACB=30 deg.,therefore LDCG=30 deg.
Since CG=CD, LCGD=LCDG=75 deg.
EF is parallel to DG, therefore LDEF=LCDG= 75 deg.
DE=EF
Therefore in triangle DEF, LE=75 and LEFD=LEDF=(180-75)/2=52.5
Answer C.
LACB=30 deg.,therefore LDCG=30 deg.
Since CG=CD, LCGD=LCDG=75 deg.
EF is parallel to DG, therefore LDEF=LCDG= 75 deg.
DE=EF
Therefore in triangle DEF, LE=75 and LEFD=LEDF=(180-75)/2=52.5
Answer C.
