The segment AS is a median of - ABC- If AB - 10- AC- 24 and AS - 13- Find BC-

In triangle ABC, Point S is the midpoint of side BC. BS=SC= 12BC By Apollonius theorem AB^2+AC^2=2AS^2+2BS^2 ⇒ (10)^2+(24)^2=2(13)^2+2(BS)^2 ⇒ 100+576=338+2( ...

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Pythagoras Theorem

The segment A...

Question

The segment AS is a median of $Δ A B C .$ If AB = 10, AC= 24 and AS = 13. Find BC.

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Solution

The correct option is C
26
In triangle ABC, Point S is the midpoint of side BC.

$B S = S C = \frac{1}{2} B C$

By Apollonius theorem
$A B^{2} + A C^{2} = 2 A S^{2} + 2 B S^{2} \Rightarrow (10)^{2} + (24)^{2} = 2 (13)^{2} + 2 (B S)^{2} \Rightarrow 100 + 576 = 338 + 2 (B S)^{2} \Rightarrow 676 = 338 + 2 (B S)^{2} \Rightarrow 676 - 338 = 2 (B S)^{2} \Rightarrow 338 = 2 (B S)^{2} \Rightarrow (B S)^{2} = 169 \Rightarrow B S = 13 S o, B C = 2 \times 13 = 26.$

Hence, option c is correct.

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The segment AS is a median of ΔABC. If AB = 10, AC= 24 and AS = 13. Find BC.

The correct option is C 26In triangle ABC, Point S is the midpoint of side BC. BS=SC=12BC By Apollonius theorem AB2+AC2=2AS2+2BS2⇒(10)2+(24)2=2(13)2+2(BS)2⇒100+576=338+2(BS)2⇒676=338+2(BS)2⇒676−338=2(BS)2⇒338=2(BS)2⇒(BS)2=169⇒BS=13So, BC=2×13=26. Hence, option c is correct.

The segment AS is a median of $Δ A B C .$ If AB = 10, AC= 24 and AS = 13. Find BC.