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In ∆Pqr, Point S is the Midpoint of Side Qr. If Pq = 11, Pr = 17, Ps =13, Find Qr. - Geometry

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Question

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS =13, find QR.

Solution

In ∆PQR, point S is the midpoint of side QR.

\[QS = SR = \frac{1}{2}QR\]

\[{PQ}^2 + {PR}^2 = 2 {PS}^2 + 2 {QS}^2 \left( \text{by Apollonius theorem} \right)\]

\[ \Rightarrow {11}^2 + {17}^2 = 2 \left( 13 \right)^2 + 2 {QS}^2 \]

\[ \Rightarrow 121 + 289 = 2\left( 169 \right) + 2 {QS}^2 \]

\[ \Rightarrow 410 = 338 + 2 {QS}^2 \]

\[ \Rightarrow 2 {QS}^2 = 410 - 338\]

\[ \Rightarrow 2 {QS}^2 = 72\]

\[ \Rightarrow {QS}^2 = 36\]

\[ \Rightarrow QS = 6\]

\[ \therefore QR = 2 \times QS\]

\[ = 2 \times 6\]

\[ = 12\]

Hence, QR = 12.

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Solution In ∆Pqr, Point S is the Midpoint of Side Qr. If Pq = 11, Pr = 17, Ps =13, Find Qr. Concept: Pythagoras Theorem.
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