Question

PQR is a triangle in which PQ=PR=26cm. S is a point on segment QR such that SR=3cm. And PS=25cm. Find length of QS.

Answer 1

$\textbf{Given:}$

$\text{In \triangle PQR, PQ=PR=26 cm, SR= cm, PS=25 cm}$

$\textbf{To find:}$

$\text{Length of QS}$

$\textbf{Solution:}$

$\text{First we find area of \triangle PSR,}$

$\text{We apply heron's formula}$

$s=\dfrac{26+25+3}{2}=\dfrac{54}{2}=27$

$\textbf{Area}\bf=\sqrt{s(s-a)(s-b)(s-c)}$

$\text{Area}=\sqrt{27(27-26)(27-25)(27-3)}$

$\text{Area}=\sqrt{27(1)(2)(24)}$

$\text{Area}=\sqrt{(9{\times}3)(2)(4{\times}3{\times}2)}$

$\text{Area}=3{\times}3{\times}2{\times}2=36\;cm^2$

$\text{Draw, PT \perp QR}$

$\text{Then,}\bf\;QT=TR$_____(1)

$\text{Now, Area of \triangle PSR}=36\;cm^2$

$\implies\,\dfrac{1}{2}{\times}SR{\times}PT=36$

$\implies\,\dfrac{1}{2}{\times}3{\times}PT=36$

$\implies\bf\,PT=24$

$\text{In right angled \triangle PTS, by pythagoras theorem}$

$PS^2=PT^2+ST^2$

$25^2=24^2+ST^2$

$625=576+ST^2$

$625-576=ST^2$

$ST^2=49$

$\implies\,ST=7\;cm$

$TR=ST+SR=7+3=10$

$\text{Using (1)}$

$\implies\;QT=10\;cm$

$\text{Finally,}$

$QS=QT+ST$

$QS=10+7$

$\implies\boxed{\bf\;QS=17\;cm}$

Find more:

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