In $Δ P Q R, P M$ is a median. If PM=26 cm and QR=24 cm, then find $P Q^{2} + P R^{2}$ .

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Solution

In $Δ$ PQR

PM is median

PM $= 26$ cm

QR $= 24$ cm

$M R = \frac{Q R}{2} = \frac{24}{2} = 12$

$Q M = M R = 12$

In $Δ$ PMR

$(P R)^{2} = (P M)^{2} + (M R)^{2}$

$(P R)^{2} = (26)^{2} + (12)^{2}$

$= 676 + 144$

$P R^{2} = 820$

$P R = 2 \sqrt{205}$

$P Q = P R = 2 \sqrt{205}$

$∴ P Q^{2} + P R^{2} = (2 \sqrt{205})^{2} + (2 \sqrt{205})^{2}$

$= 4 (205) + 4 (205)$

$= 820 + 820$

$= 1640$ .

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Similar questions

△

P M = 9

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

= 290

Q R

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

Δ P Q R

Δ P Q R,

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

Q M ⊥ P R

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

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