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In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is

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In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   11 Oct 2013, 22:50
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In triangle PQR, the angle Q = 90 degree, PQ = 6 cm, QR = 8 cm. X is a variable point on PQ. The line through X parallel to QR, intersects PR at Y and the line through Y, parallel to PQ, intersects QR at Z. Find the least possible length of XZ

A. 3.6 cm
B. 2.4 cm
C. 4.8 cm
D. 2.16 cm
E. 3.2 cm
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C
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Bunuel
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In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   12 Oct 2013, 06:23
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Qoofi wrote:
In triangle PQR, the angle Q = 90 degree, PQ = 6 cm, QR = 8 cm. X is a variable point on PQ. The line through X parallel to QR, intersects PR at Y and the line through Y, parallel to PQ, intersects QR at Z. Find the least possible length of XZ

A. 3.6 cm
B. 2.4 cm
C. 4.8 cm
D. 2.16 cm
E. 3.2 cm


Good question. +1.

Look at the diagram below:



Notice that QZYX is a rectangle, thus diagonals XZ and QY are equal. So, we need to minimize QY. QY (the distance from Q to the hypotenuse) will be minimized when QY is perpendicular to the hypotenuse.

Now, in case when QY is perpendicular to PR, two right triangles PQR and PQY are similar: QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8.

Answer: C.

Hope it's clear.

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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   23 Oct 2013, 01:06
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... another approach:

I'm gonna refer to the graphics bunuel provided:
From the information given in the question, we can compute the area of the triangle = 6*8*(1/2)=24
As bunuel said ZX and QY are equal, and QY is also PR's height of the triangle.
Since we know the triangles area and the length of PR we can compute:
24=10*x*(1/2)
x=4,8
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   20 Oct 2013, 05:33
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arakban99 wrote:
Bunuel wrote:
bluecatie1 wrote:
Hi Bunuel,

I do have a bit of a simple question regarding this problem, or rather similar triangles in particular...

When you write that QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8 then does QY 'correspond' with QR becasue they share the same angle, which is angle P?

It's very easy for me to get mixed up with those side & angle associations in similar triangles, so I just wanted to verify...

Many thanks.


Yes, you are correct: corresponding sides are opposite corresponding angles.



How do you understand what sides are corresponding, the other two angles are not given for each right angle triangle.


Right angle Q in triangle RQP corresponds to right angle Y in triangle QYP.
Angle P in triangle RQP corresponds to the same angle P in triangle QYP.
The third angles must also be equal, thus angle R in triangle RQP corresponds to angle YQP in triangle QYP.

Finally, corresponding sides are opposite corresponding angles.

Does this make sense?
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   16 Oct 2013, 05:07
Hi Bunuel,

I do have a bit of a simple question regarding this problem, or rather similar triangles in particular...

When you write that QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8 then does QY 'correspond' with QR becasue they share the same angle, which is angle P?

It's very easy for me to get mixed up with those side & angle associations in similar triangles, so I just wanted to verify...

Many thanks.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   16 Oct 2013, 11:11
Do we get such difficult questions in GMAT?
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   17 Oct 2013, 02:56
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bluecatie1 wrote:
Hi Bunuel,

I do have a bit of a simple question regarding this problem, or rather similar triangles in particular...

When you write that QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8 then does QY 'correspond' with QR becasue they share the same angle, which is angle P?

It's very easy for me to get mixed up with those side & angle associations in similar triangles, so I just wanted to verify...

Many thanks.


Yes, you are correct: corresponding sides are opposite corresponding angles.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   19 Oct 2013, 09:24
Bunuel wrote:
bluecatie1 wrote:
Hi Bunuel,

I do have a bit of a simple question regarding this problem, or rather similar triangles in particular...

When you write that QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8 then does QY 'correspond' with QR becasue they share the same angle, which is angle P?

It's very easy for me to get mixed up with those side & angle associations in similar triangles, so I just wanted to verify...

Many thanks.


Yes, you are correct: corresponding sides are opposite corresponding angles.



How do you understand what sides are corresponding, the other two angles are not given for each right angle triangle.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   20 Oct 2013, 13:02
Quote:

Does this make sense?


yes
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   22 Oct 2013, 22:41
Qoofi wrote:
wiut wrote:
I have chosen "B" , but the OA seems to be "C"


Answer is C. 4.8. Wish I could edit Bunuel's post :(


Please tell how you assumes in the first place that XYQZ is a rectangle & secondly why you have not taken triangle XYZ to prove the congruency directly?
Please reply.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   23 Oct 2013, 00:26
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anu1706 wrote:
Qoofi wrote:
wiut wrote:
I have chosen "B" , but the OA seems to be "C"


Answer is C. 4.8. Wish I could edit Bunuel's post :(


Please tell how you assumes in the first place that XYQZ is a rectangle & secondly why you have not taken triangle XYZ to prove the congruency directly?
Please reply.


We know that XY is parallel to QR, YZ is parallel to PQ and angle Q is 90 degrees --> QZYX is a rectangle.

As for your second question: triangles considered are similar not congruent. Please show which triangle is similar to XYX and why.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   23 Oct 2013, 22:53
Bunuel wrote:
Qoofi wrote:
In triangle PQR, the angle Q = 90 degree, PQ = 6 cm, QR = 8 cm. X is a variable point on PQ. The line through X parallel to QR, intersects PR at Y and the line through Y, parallel to PQ, intersects QR at Z. Find the least possible length of XZ

A. 3.6 cm
B. 2.4 cm
C. 4.8 cm
D. 2.16 cm
E. 3.2 cm


Good question. +1.

Look at the diagram below:
Attachment:
Untitled2.png
Notice that QZYX is a rectangle, thus diagonals XZ and QY are equal. So, we need to minimize QY. QY (the distance from Q to the hypotenuse) will be minimized when QY is perpendicular to the hypotenuse.

Now, in case when QY is perpendicular to PR, two right triangles PQR and PQY are similar: QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8.

Answer: C.

Hope it's clear.


Hi Bunuel,

I solved this question differently and getting option A as the answer.Can you please correct me where as to where am I going wrong?

Since the shortest length is the perpendicular length, two right angled trianles PQY and YQR are formed.Let the length PY=x & YR=10-x

Therefore,
8^2-(10-x)^2=6^2-x^2
solving x=3.6

Thanks so much!
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   24 Oct 2013, 00:55
Expert Reply
aryaparul wrote:
Bunuel wrote:
Qoofi wrote:
In triangle PQR, the angle Q = 90 degree, PQ = 6 cm, QR = 8 cm. X is a variable point on PQ. The line through X parallel to QR, intersects PR at Y and the line through Y, parallel to PQ, intersects QR at Z. Find the least possible length of XZ

A. 3.6 cm
B. 2.4 cm
C. 4.8 cm
D. 2.16 cm
E. 3.2 cm


Good question. +1.

Look at the diagram below:
Attachment:
Untitled2.png
Notice that QZYX is a rectangle, thus diagonals XZ and QY are equal. So, we need to minimize QY. QY (the distance from Q to the hypotenuse) will be minimized when QY is perpendicular to the hypotenuse.

Now, in case when QY is perpendicular to PR, two right triangles PQR and PQY are similar: QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8.

Answer: C.

Hope it's clear.


Hi Bunuel,

I solved this question differently and getting option A as the answer.Can you please correct me where as to where am I going wrong?

Since the shortest length is the perpendicular length, two right angled trianles PQY and YQR are formed.Let the length PY=x & YR=10-x

Therefore,
8^2-(10-x)^2=6^2-x^2
solving x=3.6

Thanks so much!


You denoted PY as x, we need to find QY. It will come as 4.8.

Hope it helps.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   02 May 2017, 12:26
Bunuel wrote:
bluecatie1 wrote:
Hi Bunuel,

I do have a bit of a simple question regarding this problem, or rather similar triangles in particular...

When you write that QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8 then does QY 'correspond' with QR becasue they share the same angle, which is angle P?

It's very easy for me to get mixed up with those side & angle associations in similar triangles, so I just wanted to verify...

Many thanks.


Yes, you are correct: corresponding sides are opposite corresponding angles.



Hi,
Can you please tell me where I'm going wrong when I consider PQR similar to PYQ?
P- same angle
Q & Y - rt angles
therefore 3rd angle equal?
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   03 May 2017, 03:35
Expert Reply
salonipatil wrote:
Bunuel wrote:
bluecatie1 wrote:
Hi Bunuel,

I do have a bit of a simple question regarding this problem, or rather similar triangles in particular...

When you write that QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8 then does QY 'correspond' with QR becasue they share the same angle, which is angle P?

It's very easy for me to get mixed up with those side & angle associations in similar triangles, so I just wanted to verify...

Many thanks.


Yes, you are correct: corresponding sides are opposite corresponding angles.



Hi,
Can you please tell me where I'm going wrong when I consider PQR similar to PYQ?
P- same angle
Q & Y - rt angles
therefore 3rd angle equal?


No mistake there. If Y is a right angle PQR and PYQ will be similar.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   07 May 2017, 14:32
roopika2990 wrote:
Do we get such difficult questions in GMAT?


Doubtful. However, the question contains a few key principles including that the shortest distance from a point to a line is through a perpendicular connector.
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Re: In triangle PQR, the angle Q = 9, PQ = 6 cm, QR = 8 cm. X is [#permalink] New post   26 Jul 2023, 03:10
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