In $△$ PQR, seg PM is the median. If $P M = 9$ and $P Q^{2} + P R^{2}$ $= 290$ . Find $Q R$ .

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Solution

In $△$ PQR,seg PM is the median (Given)
$P Q^{2} + P R^{2} = 2 P M^{2} + 2 Q M^{2}$ (By Appollonius theorem)
$290 = 2 (9)^{2} + 2 Q M^{2}$
$290 = 2 (81) + 2 Q M^{2}$
$290 = 162 + 2 Q M^{2}$
$290162 = 2 Q M^{2}$
$128 = 2 Q M^{2}$
$Q M^{2} = \frac{128}{2}$
$Q M^{2} = 64$
$Q M = 8$ units
$Q M = \frac{1}{2} Q R$
$Q R = 2 Q M$
$Q R = 16 u n i t s$

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Similar questions

In △ PQR, seg PM is the median . If PM = 9 and {PQ}^{2} + {PR}^{2} = 290, f ind QR .

Δ P Q R, P M

P Q^{2} + P R^{2}

△ P Q R, Q M

P R

P R^{2} - P Q^{2} = Q R^{2}

Q M^{2} = P M \times M R

∠ Q P R = 90^{\circ}, s e g P M ⊥ s e g Q R

P M = 10, Q M = 8,

Δ

P Q R

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