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In the above figure, if PQRS is a square, PT is perpendicular to QT

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In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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In the above figure, if PQRS is a square, PT is perpendicular to QT, QT = 3 and PT = 4, then RT =

(A) 5√2
(B) 2√15
(C) √58
(D) √65
(E) √66


[Reveal] Spoiler:
Attachment:
2017-11-07_0940.png
2017-11-07_0940.png [ 5.91 KiB | Viewed 887 times ]
[Reveal] Spoiler: OA

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In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 20 Nov 2017, 17:32
Bunuel wrote:
Image
In the above figure, if PQRS is a square, PT is perpendicular to QT, QT = 3 and PT = 4, then RT =

(A) 5√2
(B) 2√15
(C) √58
(D) √65
(E) √66


[Reveal] Spoiler:
Attachment:
The attachment 2017-11-07_0940.png is no longer available

Attachment:
sqinsq.png
sqinsq.png [ 28.33 KiB | Viewed 474 times ]

In the diagram, the figure on the right is to demonstrate that the leg lengths of right \(\triangle\) RTZ are 3 and 7.
Otherwise, ignore it.

I get Answer C.

To solve:

Draw a square around square PQRS, see LEFT SIDE diagram, square XYZT.

Squares are symmetrical. A symmetrical figure inscribed in a symmetrical figure will divide sides proportionally (or, if vertices are at midpoints, equally).

A square inscribed in a square will create congruent triangles where inner square vertices meet outer square sides.

The vertices of PQRS divide the sides of outside square XYZY in identical ratios: a: b, a: b, a: b, a: b

Vertices of PQRS create 4 congruent triangles whose corresponding sides are equal.

Here, a : b = 3 : 4 for all sides of square XYZT

Connect R and T. Right \(\triangle\) RTZ has
One leg length = 3 (RZ)
Other leg length = 7 (ZT)

Pythagorean theorem:
\(RT^2 = leg^2 + leg^2\)
\(RT^2 = 3^2 + 7^2\)
\(RT^2 = 9 + 49\)
\(RT^2 = 58\)
\(RT = \sqrt{58}\)

Answer C

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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 26 Nov 2017, 09:16
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amanvermagmat
Your approach to this question is not perfect, actually, the square in this question is not placed symmetrically wrt axes. Hence the diagonal is not perpendicular to PT. See the attachment. So RPT is not right-angled triangle :sad: . So further explanation is fallacious.
the right solution has been posted and the answer is C :thumbup: .

Give Kudos , if you like my post :-) .

amanvermagmat wrote:
In right angled triangle PTQ, we can get PQ = 5. Thus all sides of this square are 5, which means diagonals of this square = PR = QS = 5√2.

Now, lets just join PR. PRT is another right angled triangle right angled at P. And RT is the hypotenuese of this triangle. So, by pytahgoras theorem, RT = √(5√2)^2+(4)^2 = √66

Hence E answer

Attachments

24115413_10156965503553289_108422922_o.jpg
24115413_10156965503553289_108422922_o.jpg [ 62.04 KiB | Viewed 321 times ]


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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 26 Nov 2017, 09:38
Bunuel wrote:
Image
In the above figure, if PQRS is a square, PT is perpendicular to QT, QT = 3 and PT = 4, then RT =

(A) 5√2
(B) 2√15
(C) √58
(D) √65
(E) √66


[Reveal] Spoiler:
Attachment:
The attachment 2017-11-07_0940.png is no longer available



HI...

join R with T and R with y axis as shown in figure attached..
the sides of square are \(\sqrt{3^3+4^2}=5\)
so triangles PQT and ARQ will be similar triangles.. HYP =5 and angles are 90, x and 90-x
so AR = 3 and AQ = 4..

now in triangle ART..
AR = 3
AT = AQ+QT=4+3=7
so RT = hyp = \(\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58}\)
C
Attachments

NEW.png
NEW.png [ 7.42 KiB | Viewed 309 times ]


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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 26 Nov 2017, 10:51
Janvisahu wrote:
amanvermagmat
Your approach to this question is not perfect, actually, the square in this question is not placed symmetrically wrt axes. Hence the diagonal is not perpendicular to PT. See the attachment. So RPT is not right-angled triangle :sad: . So further explanation is fallacious.
the right solution has been posted and the answer is C :thumbup: .

Give Kudos , if you like my post :-) .

amanvermagmat wrote:
In right angled triangle PTQ, we can get PQ = 5. Thus all sides of this square are 5, which means diagonals of this square = PR = QS = 5√2.

Now, lets just join PR. PRT is another right angled triangle right angled at P. And RT is the hypotenuese of this triangle. So, by pytahgoras theorem, RT = √(5√2)^2+(4)^2 = √66

Hence E answer


Hi

Thanks for bringing it to my attention. I have deleted that post in the light of this new knowledge.

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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 26 Nov 2017, 11:48
chetan2u wrote:
Bunuel wrote:
Image
In the above figure, if PQRS is a square, PT is perpendicular to QT, QT = 3 and PT = 4, then RT =

(A) 5√2
(B) 2√15
(C) √58
(D) √65
(E) √66


[Reveal] Spoiler:
Attachment:
2017-11-07_0940.png



HI...

join R with T and R with y axis as shown in figure attached..
the sides of square are \(\sqrt{3^3+4^2}=5\)
so triangles PQT and ARQ will be similar triangles.. HYP =5 and angles are 90, x and 90-x
so AR = 3 and AQ = 4..

now in triangle ART..
AR = 3
AT = AQ+QT=4+3=7
so RT = hyp = \(\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58}\)
C

Hi Chetan,

In your solution why PT is not equal to AR.. As PQRS is a square I think PT = AR. Can you please help me to clarify above doubt

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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 26 Nov 2017, 15:49
rahul16singh28 wrote:
chetan2u wrote:
Bunuel wrote:
Image
In the above figure, if PQRS is a square, PT is perpendicular to QT, QT = 3 and PT = 4, then RT =

(A) 5√2
(B) 2√15
(C) √58
(D) √65
(E) √66


[Reveal] Spoiler:
Attachment:
2017-11-07_0940.png



HI...

join R with T and R with y axis as shown in figure attached..
the sides of square are \(\sqrt{3^3+4^2}=5\)
so triangles PQT and ARQ will be similar triangles.. HYP =5 and angles are 90, x and 90-x
so AR = 3 and AQ = 4..

now in triangle ART..
AR = 3
AT = AQ+QT=4+3=7
so RT = hyp = \(\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58}\)
C

Hi Chetan,

In your solution why PT is not equal to AR.. As PQRS is a square I think PT = AR. Can you please help me to clarify above doubt

Sent from my Lenovo P1a42 using GMAT Club Forum mobile app


The figure it not drawn accurately and thus causes confusion. PT = 4 and TQ = 3, but they both look like they equal 4. Because of this, it seems like the square is just turned 45 degrees and you can draw a perpendicular line down from PT to PR - which is in fact, not the case. If you were to create an accurate drawing the line would be skewed.

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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT [#permalink]

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New post 27 Nov 2017, 12:09
Bunuel wrote:
Image
In the above figure, if PQRS is a square, PT is perpendicular to QT, QT = 3 and PT = 4, then RT =

(A) 5√2
(B) 2√15
(C) √58
(D) √65
(E) √66


[Reveal] Spoiler:
Attachment:
2017-11-07_0940.png


We see that triangle QPT is a 3-4-5 right triangle; thus PQ = 5. The way the figure is drawn, we are being tricked into thinking that the diagonal of the square is perpendicular to the line PT, but we should notice that that would happen only when the angle TPQ is 45 degrees, which is not the case since the triangle PTQ is not isosceles.

Instead, draw segments RT and drop a perpendicular from R into line QT. Let’s say the perpendicular meets the line QT at the point W. Now, we have a right triangle where the hypotenuse is RT and the legs are RW and TW. Since the triangle PTQ is similar to the triangle RQW (notice that the angle PQT is equal to the angle QRW), the length of the segment QW will be 4 and the length of the segment RW will be 3. Now, the two legs of the right triangle RTW are 3 and 3 + 4 = 7, and the hypotenuse of RTW, which is RT, can be found using the Pythagorean Theorem: |RT|^2 = 3^2 + 7^2 = 9 + 49 = 58; thus RT = √58.

Answer: C
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Re: In the above figure, if PQRS is a square, PT is perpendicular to QT   [#permalink] 27 Nov 2017, 12:09
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