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.
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