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.
In the triangle shown above, BA is parallel to CE. What is t
[#permalink]
19 Sep 2017, 09:49
19 Sep 2017, 09:49
1
Expert Reply
00:00
Question Stats:
97% (00:59) correct
2% (03:16) wrong based on 44 sessions
Hide Show timer Statistics
Attachment:
triangle.jpg [ 21.75 KiB | Viewed 4364 times ]
In the triangle shown above, BA is parallel to CE. What is the value of x + y?
enter your value
Show: :: OA
190
Re: In the triangle shown above, BA is parallel to CE. What is t
[#permalink]
25 Sep 2017, 04:55
25 Sep 2017, 04:55
1
Considering the trapezoid AECB, we have that both angles A and B are 90° so that y° = 360-90-90-40 = 140. Then, considering the triangle CDE, angle E is 90 degrees, angle C is 180-y = 40 and then x = 180-90-40 = 50.
Thus, x+y = 190!
Thus, x+y = 190!
Re: In the triangle shown above, BA is parallel to CE. What is t
[#permalink]
25 Sep 2017, 05:20
25 Sep 2017, 05:20
1
Carcass wrote:
In the triangle shown above, BA is parallel to CE. What is the value of x + y?
enter your value
Show: :: OA
190
SInce CE parallel to BA
therefore we have angle B = angle Z = 40 degree (plz refer the diagram)
and angle A = angle E = 90 degree
NOw,
in triangle CED we have,
the sum of all angles = 180 degree
Therefore x+90 + 40 = 180 degree
or x = 50 degree.
Since angle BCD = 180 (since the two angles meet at C make up a straight line and they sum up to 180 degree)
Therefore angle y + angle Z = 180
or angle y = 180 -40 = 140 degree.
Now x+y = 140 +50 = 190 degree.
Attachments
triangle.jpg [ 30.31 KiB | Viewed 4154 times ]
