Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM

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Solution

In ∆ABC, point M is the midpoint of side BC.

$BM = MC = \frac{1}{2} BC = 12$

${AB}^{2} + {AC}^{2} = 2 {AM}^{2} + 2 {BM}^{2} (by Apollonius theorem) \Rightarrow {(22)}^{2} + {(34)}^{2} = 2 {AM}^{2} + 2 {(12)}^{2} \Rightarrow 484 + 1156 = 2 {AM}^{2} + 288 \Rightarrow 1640 - 288 = 2 {AM}^{2} \Rightarrow 1352 = 2 {AM}^{2} \Rightarrow {AM}^{2} = 676 \Rightarrow AM = 26$

Hence, AM = 26.

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Similar questions

Seg AD is the median of △ ABC and AM ⊥ BC . Prove that (i) {AC}^{2} = {AD}^{2} + BC \times DM + {(\frac{BC}{2})}^{2} (ii) {AB}^{2} = {AD}^{2} + BC \times DM + {(\frac{BC}{2})}^{2}

{AB}^{2} = {AD}^{2} - BC \times DM + {(\frac{BC}{2})}^{2}

Δ A B C .

AM is a median of ∆ABC. Prove that (AB + BC + CA) > 2AM.

Δ A B C

A D

A B = 11, A C = 17, B C = 12

A D

If AM is a median of triangle ABC. Then AB +BC +CA $>$ 2AM

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