In ∆ABC, seg AP is a median. If BC = 18, AB2 + AC2 = 260, Find AP. - Geometry Mathematics 2

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Sum

In ∆ABC, seg AP is a median. If BC = 18, AB2 + AC2 = 260, Find AP.

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Solution

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

\[BP = PC = \frac{1}{2}BC = 9\]
\[{AB}^2 + {AC}^2 = 2 {AP}^2 + 2 {BP}^2\] ...(by Apollonius theorem)
\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 9^2 \right)\]
\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 81 \right)\]
\[ \Rightarrow 260 = 2 {AP}^2 + 162\]
\[ \Rightarrow 2 {AP}^2 = 260 - 162\]
\[ \Rightarrow 2 {AP}^2 = 98\]
\[ \Rightarrow {AP}^2 = 49\]
\[ \Rightarrow AP = 7\]

Hence, AP = 7.

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Chapter 2: Pythagoras Theorem - Problem Set 2 [Page 44]

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Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Problem Set 2 | Q 6 | Page 44

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