Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 18 | Page 46
Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM
solution
[BM = MC = frac{1}{2}BC = 12]
[{AB}^2 + {AC}^2 = 2 {AM}^2 + 2 {BM}^2 left( text{by Apollonius theorem} right)]
[ Rightarrow left( 22 right)^2 + left( 34 right)^2 = 2 {AM}^2 + 2 left( 12 right)^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.
Explanation:-
Given: $BM = MC = frac{1}{2}BC = 12$
Using the Apollonius theorem in triangle $ABC$ with respect to median $AM$,
begin{align*} {AB}^2 + {AC}^2 &= 2 {AM}^2 + 2 {BM}^2 Rightarrow (22)^2 + (34)^2 &= 2 {AM}^2 + 2 (12)^2 Rightarrow 484 + 1156 &= 2 {AM}^2 + 288 Rightarrow 1640 &= 2 {AM}^2 Rightarrow {AM}^2 &= 820 Rightarrow AM &= sqrt{820} = 2sqrt{205} end{align*}
Hence, $AM = 2sqrt{205}$.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 18 | Page 46