Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 17 | Page 44
Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR
solution
In ∆PQR, point M is the midpoint of side QR.
[QM = MR = frac{1}{2}QR]
[{PQ}^2 + {PR}^2 = 2 {PM}^2 + 2 {QM}^2 left( text{by Apollonius theorem} right)]
[ Rightarrow left( 40 right)^2 + left( 42 right)^2 = 2 left( 29 right)^2 + 2 {QM}^2 ]
[ Rightarrow 1600 + 1764 = 1682 + 2 {QM}^2 ]
[ Rightarrow 3364 – 1682 = 2 {QM}^2 ]
[ Rightarrow 1682 = 2 {QM}^2 ]
[ Rightarrow {QM}^2 = 841]
[ Rightarrow QM = 29]
[ Rightarrow QR = 2 times 29]
[ Rightarrow QR = 58]
Hence, QR = 58.
Explanation:-
In ∆PQR, point M is the midpoint of side QR. QM = MR = (1/2)QR
By Apollonius theorem, PQ^2 + PR^2 = 2PM^2 + 2QM^2
=> (40)^2 + (42)^2 = 2(29)^2 + 2QM^2 => 1600 + 1764 = 1682 + 2QM^2 => 1682 = 2QM^2 => QM^2 = 841 => QM = 29 => QR = 2×29 => QR = 58
Hence, QR = 58.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 17 | Page 46