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In the figure, AD and BC are lines intersecting at O. What i
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22 Sep 2018, 09:21
22 Sep 2018, 09:21
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Expert Reply
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Question Stats:
91% (01:46) correct
8% (01:40) wrong based on 37 sessions
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#GREpracticequestion In the figure, AD and BC are lines intersecting at O.jpg [ 20.85 KiB | Viewed 4585 times ]
In the figure, AD and BC are lines intersecting at O. What is the value of a ?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 135
Re: In the figure, AD and BC are lines intersecting at O. What i
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01 Jul 2020, 22:50
01 Jul 2020, 22:50
1
To solve this problem, we should work with opposite angles which values are the same.
This means that:
5y/2=5x+5a
Reducing
y/2=x+a
a=y/2-x............................(1)
Also
y=2x+30
Reducing this expression, we obtain
y/2=x+15 .........................(2)
Replacing (2) in (1)
a=(x+15)-x
a=15
Answer=A
This means that:
5y/2=5x+5a
Reducing
y/2=x+a
a=y/2-x............................(1)
Also
y=2x+30
Reducing this expression, we obtain
y/2=x+15 .........................(2)
Replacing (2) in (1)
a=(x+15)-x
a=15
Answer=A
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Re: In the figure, AD and BC are lines intersecting at O. What i
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13 Feb 2021, 01:34
13 Feb 2021, 01:34
1
Carcass wrote:
Attachment:
#GREpracticequestion In the figure, AD and BC are lines intersecting at O.jpg
In the figure, AD and BC are lines intersecting at O. What is the value of a ?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 135
Vertically opposite angles are equal
So, \(y = 2x + 30\)
and, \(5x + 5a\) = \(\frac{5(2x +30)}{2}\)
\(10x + 10a\) =\(10x + 150\)
\(10a = 150\)
\(a = 15\)
Hence, option A
