Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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In ∆Abc Seg Ap is a Median. If Bc = 18, Ab2 + Ac2 = 260 Find Ap. - Geometry

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Question

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

Solution

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

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

Hence, AP = 7.

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APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Problem Set 2 | Q: 6 | Page no. 44
Solution In ∆Abc Seg Ap is a Median. If Bc = 18, Ab2 + Ac2 = 260 Find Ap. Concept: Apollonius Theorem.
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