Problem: The sides PQ and PR of triangle PQR are respectively of lengths 4 inches and 7 inches. The median PM is 321 inches. Then QR, in inches, is:
Answer Choices:
A. 6
B. 7
C. 8
D. 9
E. 10
Solution:
16−(y−x)2=(27)2−x2∴(y−x)2−x2=415
16−(y−x)2=72−(y+x)2∴(y+x)2−(y−x)2=33
∴y2−2xy=415 and 2xy=233∴y2=481,y=29∴QR=9
By the law of cosines, (27)2+y2−2⋅27⋅ycos∠PMQ=16 and (27)2+y2−2⋅27⋅ycos(180∘−∠PMQ) =(27)2+y2+7ycos∠PMQ=49∴249+2y2=65,y=29,2y=9
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