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PQRS is a parallelogram and ST = TR. What is the ratio of the area of

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Bunuel
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PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink] New post   16 May 2016, 02:33
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PQRS is a parallelogram and ST = TR. What is the ratio of the area of triangle QST to the area of the parallelogram?

A. 1 : 2
B. 1 : 3
C. 1 : 4
D. 1 : 5
E. it cannot be determined

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BrentGMATPrepNow
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Re: PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink] New post   07 Sep 2018, 15:09
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Bunuel wrote:

PQRS is a parallelogram and ST = TR. What is the ratio of the area of triangle QST to the area of the parallelogram?

A. 1 : 2
B. 1 : 3
C. 1 : 4
D. 1 : 5
E. it cannot be determined

Show Spoiler
Attachment:
2016-05-16_1331.png


Since ST = TR, let's let x = each length


Also, let's say the parallelogram has height h


So, the length of the parallelogram's base = x + x = 2x
Area of a PARALLELOGRAM = (base)(height)
= (2x)(h)
= 2xh


Now let's determine the area of triangle QST

Area of any triangle = (base)(height)/2
So, area of triangle QST = xh/2

What is the ratio of the area of triangle QST to the area of the parallelogram?
Answer = xh/2 : 2xh

Looks like we need to simplify our ratio.
Take: xh/2 : 2xh
Divide both sides by xh to get: 1/2 : 2_
Multiply both sides by 2 to get: 1 : 4

Answer: C

Cheers,
Brent
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Re: PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink] New post   16 May 2016, 06:20
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Bunuel wrote:

PQRS is a parallelogram and ST = TR. What is the ratio of the area of triangle QST to the area of the parallelogram?

A. 1 : 2
B. 1 : 3
C. 1 : 4
D. 1 : 5
E. it cannot be determined

Show Spoiler
Attachment:
2016-05-16_1331.png


The diagonal of llgm divides the AREA in two equal parts..
so SQ divides the area in two equal parts ..
AREA of QRS = 1/2 of PQRS..

Now in QSR, QT divides the triangle QST in two equal areas,
Because the height of QST ans QRT is SAME and it is given that the BASE in both cases ST and TR are also same ...
SO Area of QST = Area of RST...
so AREA of QST = \(\frac{1}{2}\)of QST = \(\frac{1}{2}\)of \(\frac{1}{2}\)of llgm PQRS \(= \frac{1}{4}\)
ans C
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Re: PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink] New post   16 May 2016, 10:30
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Bunuel wrote:

PQRS is a parallelogram and ST = TR. What is the ratio of the area of triangle QST to the area of the parallelogram?

A. 1 : 2
B. 1 : 3
C. 1 : 4
D. 1 : 5
E. it cannot be determined

Show Spoiler
Attachment:
2016-05-16_1331.png


Given ST = TR ; so T must be the midpoint of SR

Area of Δ SQT = Δ QTR

Further ; Area of Δ PSQ = Area of Δ QSR ( Since the parallelogram is made up of 2 equal triangles)

So we may say that the parallelogram is made up of the area of 4 equal Δs of area QST

So, the //gm PQRS = 4 Δ QST

Hence the ratio of the area of triangle QST to the area of the parallelogram = 1 : 4 ; answer will be (C)
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Re: PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink] New post   16 May 2016, 13:15
Area of the parallelogram is = perpendicular * base (SR)
Area of the triangle is = (1/2) * perpendicular * base (ST) = (1/2) * perpendicular * SR/2

Ratio = 1/4. C

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Re: PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink] New post   05 Feb 2023, 16:52
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Re: PQRS is a parallelogram and ST = TR. What is the ratio of the area of [#permalink]
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