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प्रश्न
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
योग
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उत्तर
In ∆PQR, point S is the midpoint of side QR.
\[QS = SR = \frac{1}{2}QR\]
\[{PQ}^2 + {PR}^2 = 2 {PS}^2 + 2 {QS}^2\] .......…[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.
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