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

Question

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

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

Bunuel · Accepted Answer

Good question. +1. Look at the diagram below: https://gmatclub.com/forum/download/file.php?id=21336 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. Untitled2.png

vogelleblanc · Answer

... 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

bluecatie1 · Answer

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.

roopika2990 · Answer

Do we get such difficult questions in GMAT?

Bunuel · Answer

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

arakban99 · Answer

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

Bunuel · Answer

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?

arakban99 · Answer

yes

anu1706 · Answer

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.

Bunuel · Answer

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.

aryaparul · Answer

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!

Bunuel · Answer

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

Hope it helps.

salonipatil · Answer

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?

Bunuel · Answer

No mistake there. If Y is a right angle PQR and PYQ will be similar.

Cez005 · Answer

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.

bumpbot · Answer

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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

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