Question

In the given figure, PQR is a triangle and S is a point on QR such that the line PS divides QR in the ratio 2 : 1. The bisectors of angles PSQ and PSR meet PQ and PR at T and U respectively. If PU = 128 cm, UR = 16 cm, and PQ = 100 cm, then PT is



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Answer 1

$\textbf{Angle bisector theorem:}$

$\text{When vertical angle of a triangle is bisected,}$

$\text{the bisector divides the base into two segments}$

$\text{which have the ratio as the order of other two sides}$

$\textbf{Given:}$

$\text{PU=128 cm, UR=16 cm, PQ=100 cm}$

$\textbf{To find: PT}$

$\textbf{Solution:}$

$\text{The point S divides QR in the ratio 2:1}$

$\text{Then,}\;QS=\frac{2}{3}QR\;\text{and}\;SR=\frac{1}{3}QR$

$\textbf{ST is the angle bisector of \angle{PSQ} }$

$\text{By Angle bisector theorem, we get}$

$\dfrac{PT}{QT}=\dfrac{PS}{QS}$

$\dfrac{x}{100-x}=\dfrac{PS}{\frac{2}{3}QR}$

$\dfrac{2x}{300-3x}=\dfrac{PS}{QR}$.........(1)

$\textbf{SU is the angle bisector of \angle{PSR} }$

$\text{By Angle bisector theorem, we get}$

$\dfrac{UR}{UP}=\dfrac{SR}{PS}$

$\dfrac{16}{128}=\dfrac{\frac{1}{3}QR}{PS}$

$\dfrac{1}{8}=\dfrac{\frac{1}{3}QR}{PS}$

$\dfrac{3}{8}=\dfrac{QR}{PS}$

$\dfrac{8}{3}=\dfrac{PS}{QR}$...(2)

$\text{From (1) and (2), we get}$

$\dfrac{2x}{300-3x}=\dfrac{8}{3}$

$\dfrac{x}{300-3x}=\dfrac{4}{3}$

$\implies\,3x=1200-12x$

$\implies\,15\,x=1200$

$\implies\,x=\dfrac{1200}{15}$

$\implies\,x=80\;\text{cm}$

$\therefore\textbf{PT is 80 cm}$

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