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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. In parallelogram PQRS, TR bisects ∠QRS.  Question banks Downloads My Bookmarks Reviews Important topics
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TAGS: Intern Joined: 07 Jan 2017
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In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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Question Stats: 77% (01:28) correct 22% (01:04) wrong based on 31 sessions 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
[Reveal] Spoiler: OA

Last edited by Carcass on 11 Feb 2017, 11:53, edited 1 time in total.
Edited by carcass GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Kudos [?]: 1910  , given: 397

Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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Expert's post
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.
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Sandy
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Director Joined: 09 Nov 2018
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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.

Please explain ∠QRT + ∠RTQ + ∠QRS= which triangle? GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 120

Kudos [?]: 1910  , given: 397

Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
1
KUDOS
Expert's post
AE wrote:

Please explain ∠QRT + ∠RTQ + ∠QRS= which triangle?

Please explain ∠QRT + ∠RTQ + ∠TQR= triangle TQR.. There was a typo fixed it!
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Sandy
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
1
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as pqrs is parallelogram pq|| sr ---- 1
thus angle(p) = angle(r)
thus r = a
thus angle(qrt) = 1/2*angle(r) = a/2
thus angle(qtr) = angle(trs)
b = a/2
2b = a therefore ans is c
Director Joined: 09 Nov 2018
Posts: 508
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Kudos [?]: 27 , given: 1

Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
sandy wrote:
AE wrote:

Please explain ∠QRT + ∠RTQ + ∠QRS= which triangle?

Please explain ∠QRT + ∠RTQ + ∠TQR= triangle TQR.. There was a typo fixed it!

Oh! thanks. I was confused.
Director Joined: 09 Nov 2018
Posts: 508
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Kudos [?]: 27 , given: 1

Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
babanya wrote:
as pqrs is parallelogram pq|| sr ---- 1
thus angle(p) = angle(r)
thus r = a
thus angle(qrt) = 1/2*angle(r) = a/2
thus angle(qtr) = angle(trs)
b = a/2
2b = a therefore ans is c

Great idea. Re: In parallelogram PQRS, TR bisects ∠QRS.   [#permalink] 11 Jan 2019, 18:34
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