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Question

# In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that (i) PT = PS, (ii) ∠PSR = 15°.

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Solution

## Given: Triangle PQR is an equilateral triangle, i.e., each angle is $60°$. QRST is a square, i.e., each angle is $90°$. $\angle PQT=\angle PQR+\angle TQR=60+90=150°$ Similarly, $\angle PRS=150°$ Consider the triangles PQT and PRS. $\mathrm{PQ}=\mathrm{PR} \left(\mathrm{Sides}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\right)\phantom{\rule{0ex}{0ex}}\mathrm{QT}=\mathrm{RS}\left(\mathrm{Sides}\mathrm{of}\mathrm{a}\mathrm{square}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{PQT}=\angle \mathrm{PRS}=150°\left(\mathrm{Proved}\right)\phantom{\rule{0ex}{0ex}}△\mathrm{PQT}\cong △\mathrm{PRS}\left(\mathrm{SAS}\mathrm{criterion}\right)\phantom{\rule{0ex}{0ex}}\mathrm{PT}=\mathrm{PS}\left(\mathrm{CPCT}\right)$ Further, the side of the square coincides with the triangle, PR = RS, which makes PRS isosceles. Let $\angle PSR=\angle SPR=x$. Using angle sum property of a triangle, we get: $150°+x+x=150°+2x=180°\phantom{\rule{0ex}{0ex}}⇒2x=30°\phantom{\rule{0ex}{0ex}}⇒x=15°$ ∴ $\angle PSR=15°$ Hence, proved.

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