Problem of the Week
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Problem E and Solution
Enough Information?
Problem
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In 4P QR, P Q = 7, P R = 9 and median P M = 7.
Determine the length of QR.
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Solution
Solution 1
Since P Q = P M = 7, 4P QM is isosceles. In 4P QM , draw an altitude from P
to QM , intersecting at N . In an isosceles triangle, the altitude drawn to the base
bisects the base. Therefore QN = N M = x. Since P M is a median in 4P QR,
M R = QM = 2x. Let P N = h.
4P N M is a right triangle. Using Pythagoras’ Theorem,
P N2 = P M2 − NM2
h2 = 72 − x2
h2 = 49 − x2
(1)
4P N R is a right triangle. Using Pythagoras’ Theorem,
P N 2 = P R2 − N R2
h2 = 92 − (x + 2x)2
h2 = 81 − (3x)2
h2 = 81 − 9x2
(2)
In equations (1) and (2), the left side of each equation is h2 . Therefore, the right
side of equation (1) must equal the right side of equation (2). So
49 − x2
−x2 + 9x2
8x2
x2
x
=
=
=
=
=
81 − 9x2
81 − 49
32
4
2, x > 0
∴ QR = QN + N M + M R = x + x + 2x = 4x = 8 units.
A second solution involving trigonometry and a system of equations is
on the next page.
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Solution 2
This solution is presented for students who have done
some trigonometry and know the law of cosines.
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Since P M is a median, QM = M R = x.
Then QR = 2x.
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Using the law of cosines in 4P QM ,
P M2
72
49
14xcos(Q)
=
=
=
=
P Q2 + QM 2 − 2(P Q)(QM )cos(Q)
72 + x2 − 2(7)(x)cos(Q)
49 + x2 − 14xcos(Q)
x2
(1)
Using the law of cosines in 4P QR,
P R2
92
81
28xcos(Q)
=
=
=
=
P Q2 + QR2 − 2(P Q)(QR)cos(Q)
72 + (2x)2 − 2(7)(2x)cos(Q)
49 + 4x2 − 28xcos(Q)
4x2 − 32
(2)
Using elimination to solve for x2 ,
(1) × 2
(2)
Subtracting
∴ the length of QR is 8 units.
28xcos(Q)
28xcos(Q)
0
2x2
x2
x
QR = 2x
=
=
=
=
=
=
=
2x2
4x2 − 32
−2x2 + 32
32
16
4, x > 0
8
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