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In parallelogram PQRS, TR bisects ∠QRS.
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Updated on: 11 Feb 2017, 11:53
Updated on: 11 Feb 2017, 11:53
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In parallelogram PQRS, TR bisects ∠QRS.
Quantity A |
Quantity B |
a |
2b |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given
Originally posted by darkdevil8z on 11 Feb 2017, 07:24.
Last edited by Carcass on 11 Feb 2017, 11:53, edited 1 time in total.
Last edited by Carcass on 11 Feb 2017, 11:53, edited 1 time in total.
Edited by carcass
Re: In parallelogram PQRS, TR bisects ∠QRS.
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12 Feb 2017, 06:05
12 Feb 2017, 06:05
3
Expert Reply
Explanation
Now we Try to find out the values of all angles in the triangle QRT.
Now TR bisects ∠QRS. and from properties of parallelogram we know that ∠QRS = a.
So ∠QRT = \(\frac{a}{2}\).
And ∠RTQ = b and finally ∠TQR= 180 - a.
Now sum of all the angles in triangle is 180.
So, ∠QRT + ∠RTQ + ∠TQR = \(\frac{a}{2} + b + 180 - a = 180\).
Simplifying \(\frac{-a}{2}+b =0\).
So a = 2b.
Hence the quantities are equal. Hence option C is correct.
Now we Try to find out the values of all angles in the triangle QRT.
Now TR bisects ∠QRS. and from properties of parallelogram we know that ∠QRS = a.
So ∠QRT = \(\frac{a}{2}\).
And ∠RTQ = b and finally ∠TQR= 180 - a.
Now sum of all the angles in triangle is 180.
So, ∠QRT + ∠RTQ + ∠TQR = \(\frac{a}{2} + b + 180 - a = 180\).
Simplifying \(\frac{-a}{2}+b =0\).
So a = 2b.
Hence the quantities are equal. Hence option C is correct.
Re: In parallelogram PQRS, TR bisects ∠QRS.
[#permalink]
14 Nov 2018, 10:53
14 Nov 2018, 10:53
sandy wrote:
Explanation
Now we Try to find out the values of all angles in the triangle QRT.
Now TR bisects ∠QRS. and from properties of parallelogram we know that ∠QRS = a.
So ∠QRT = \(\frac{a}{2}\).
And ∠RTQ = b and finally ∠QRS= 180 - a.
Now sum of all the angles in triangle is 180.
So, ∠QRT + ∠RTQ + ∠QRS = \(\frac{a}{2} + b + 180 - a = 180\).
Simplifying \(\frac{-a}{2}+b =0\).
So a = 2b.
Hence the quantities are equal. Hence option C is correct.
Now we Try to find out the values of all angles in the triangle QRT.
Now TR bisects ∠QRS. and from properties of parallelogram we know that ∠QRS = a.
So ∠QRT = \(\frac{a}{2}\).
And ∠RTQ = b and finally ∠QRS= 180 - a.
Now sum of all the angles in triangle is 180.
So, ∠QRT + ∠RTQ + ∠QRS = \(\frac{a}{2} + b + 180 - a = 180\).
Simplifying \(\frac{-a}{2}+b =0\).
So a = 2b.
Hence the quantities are equal. Hence option C is correct.
Please explain ∠QRT + ∠RTQ + ∠QRS= which triangle?
