GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
F is the midpoint of AE, and D is the midpoint of CE. Wh
[#permalink]
02 Aug 2017, 08:37
02 Aug 2017, 08:37
1
Expert Reply
12
Bookmarks
00:00
Question Stats:
18% (02:30) correct
81% (02:23) wrong based on 116 sessions
Hide Show timer Statistics
Attachment:
#GREpracticequestion F is the midpoint of AE, and D is the midpoint of CE.jpg [ 20.98 KiB | Viewed 9423 times ]
F is the midpoint of AE, and D is the midpoint of CE.
Which of the following statements MUST be true?
Indicate all such statements.
❑ FD is parallel to AC.
❑ The area of triangle DEF equals the area of triangle CDF.
❑ The area of triangle ABF is less than the area of triangle DEF.
❑ Angle AFB equals angle BFC.
❑ The area of triangle ACE equals AC x BF.
Re: F is the midpoint of AE, and D is the midpoint of CE. Wh
[#permalink]
24 Sep 2017, 08:34
24 Sep 2017, 08:34
1
Any hint for this question?
Re: F is the midpoint of AE, and D is the midpoint of CE. Wh
[#permalink]
25 Sep 2017, 02:34
25 Sep 2017, 02:34
3
Expert Reply
A is right
Because DE is \(\frac{1}{2}\) of CE and FE is \(\frac{1}{2}\) of AE (i.e., corresponding sides have proportional lengths), and angle DEF is shared between /the two triangles, you can see that triangles ACE and FDE must be similar. Similar triangles have the further property that corresponding angles are of equal measure. Thus, for example, angle DFE equals angle CAE, and so FD is parallel to AC.
B is right
Looking at the two smaller triangles in the right half of the figure, you can see that triangle CD F and triangle DEF have collinear and equal “bases” (CD and DE), and share their third vertex (F), which is some fixed distance away from
CE. Because CE is comprised of bases CD and DE of triangles CD F and DEF, respectively, the two triangles have the same height. Because CDF and DEF have equal bases and the same height, they must have the same area. (For a similar reason, triangles ACF and FCE must have equal areas; more on that later.)
Choice E is right
Because triangles ACF and FCE must have equal areas as indicated above, you can see that the area of triangle ACE must be twice that of triangle ACF. Note that FD is parallel to AC due to (true) choice (A), and BF is perpendicular toFD. Therefore, BF must be perpendicular to AC as well. Put differently, AC can be regarded as the base, and BF the height, of triangle ACF. The area of triangle ACF equals ^ times AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF.
Because DE is \(\frac{1}{2}\) of CE and FE is \(\frac{1}{2}\) of AE (i.e., corresponding sides have proportional lengths), and angle DEF is shared between /the two triangles, you can see that triangles ACE and FDE must be similar. Similar triangles have the further property that corresponding angles are of equal measure. Thus, for example, angle DFE equals angle CAE, and so FD is parallel to AC.
B is right
Looking at the two smaller triangles in the right half of the figure, you can see that triangle CD F and triangle DEF have collinear and equal “bases” (CD and DE), and share their third vertex (F), which is some fixed distance away from
CE. Because CE is comprised of bases CD and DE of triangles CD F and DEF, respectively, the two triangles have the same height. Because CDF and DEF have equal bases and the same height, they must have the same area. (For a similar reason, triangles ACF and FCE must have equal areas; more on that later.)
Choice E is right
Because triangles ACF and FCE must have equal areas as indicated above, you can see that the area of triangle ACE must be twice that of triangle ACF. Note that FD is parallel to AC due to (true) choice (A), and BF is perpendicular toFD. Therefore, BF must be perpendicular to AC as well. Put differently, AC can be regarded as the base, and BF the height, of triangle ACF. The area of triangle ACF equals ^ times AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF.
