Practice Set 2.2 | Q 1 | Page 43
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
In ∆PQR, point S is the midpoint of side QR.
\[{PQ}^2 + {PR}^2 = 2 {PS}^2 + 2 {QS}^2\] .......…[Apollonius theorem]
\[ \Rightarrow {11}^2 + {17}^2 = 2 \left( 13 \right)^2 + 2 {QS}^2 \]
\[ \Rightarrow 121 + 289 = 2\left( 169 \right) + 2 {QS}^2 \]
\[ \Rightarrow 410 = 338 + 2 {QS}^2 \]
\[ \Rightarrow 2 {QS}^2 = 410 - 338\]
\[ \Rightarrow 2 {QS}^2 = 72\]
\[ \Rightarrow {QS}^2 = 36\]
\[ \Rightarrow QS = 6\]
\[ \therefore QR = 2 \times QS\]
\[ = 2 \times 6\]
\[ = 12\]
Hence, QR = 12.
Explanation:-
We are given a triangle ∆PQR, where point S is the midpoint of side QR. Thus, we can say that:
QS = SR = 1/2 QR
Using the Apollonius theorem, we can write:
PQ^2 + PR^2 = 2PS^2 + 2QS^2
Substituting the given values, we get:
11^2 + 17^2 = 2(13^2) + 2(QS^2)
Simplifying the above equation, we get:
410 = 338 + 2(QS^2)
2(QS^2) = 410 – 338
2(QS^2) = 72
QS^2 = 36
Taking the square root on both sides, we get:
QS = 6
Since QS = SR and QR = 2QS, we can say that:
QR = 2 x QS = 2 x 6 = 12
Therefore, the length of QR is 12.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.2 | Q 1 | Page 43