(C) Given,in $\triangle PQR$,$PQ=3$Altitude $RS=\sqrt{3}$$PS=QR$In $\triangle SQR$,by the Pythagorean theorem,$QR^2=SR^2+SQ^2$.Since $PS=QR$,we have $

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(C) Given,in $\triangle PQR$,$PQ=3$Altitude $RS=\sqrt{3}$$PS=QR$In $\triangle SQR$,by the Pythagorean theorem,$QR^2=SR^2+SQ^2$.Since $PS=QR$,we have $PS^2=SR^2+SQ^2$.We know $SQ=PQ-PS=3-PS$.Substituting the values,$PS^2=(\sqrt{3})^2+(3-PS)^2$.$PS^2=3+9-6PS+PS^2$$6PS=12 \Rightarrow PS=2$.In $\triangle PRS$,by the Pythagorean theorem,$PR^2=PS^2+RS^2$.$PR^2=2^2+(\sqrt{3})^2=4+3=7$.Therefore,$PR=\sqrt{7}$.

Class 11 Mathematics Straight Line Problems related to triangle and quadrilateral

Let $PQR$ be a triangle in which $PQ=3$. From the vertex $R$,draw the altitude $RS$ to meet $PQ$ at $S$. Assume that $RS=\sqrt{3}$ and $PS=QR$. Then,$PR$ equals

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$\sqrt{5}$
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$\sqrt{6}$
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$\sqrt{7}$
D
$\sqrt{8}$

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Let $PQR$ be a triangle in which $PQ=3$. From the vertex $R$,draw the altitude $RS$ to meet $PQ$ at $S$. Assume that $RS=\sqrt{3}$ and $PS=QR$. Then,$PR$ equals

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