PQRS is a square and SRT is an equilateral triangle. prove that PT=QT and TQR = 15 degree
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Solution:-
Since PQRS is a square,
angle PSR = angle QRS... (each 90°)
Now, again in equilateral triangle SRT,
we have angle TSR = angle TRS ...(each 60°)
Angle PSR + Angle TSR = Angle QRS + Angle TRS
⇒ Angle TSP = Angle TRQ
Now in Δ TSP and Δ TRQ, we have
TS = TR (Sides of equilateral triangle)
Angle TSP = Angle TRQ... (Proved above)
PS = QR ...(Sides of square)
Δ TSP ≡ Δ TRQ
So, PT = QT Proved.
Now in Δ TQR, we have
TR = QR ....(QR = RS = TR)
Angle TQR = Angle QTR and Angle TQR + Angle QTR + Angle TRQ = 180°
Angle TQR + Angle QTR + Angle TRS + Angle SRQ = 180°
2(Angle TQR) + 60° + 90° = 180° .....(∴ ∠ TQR = ∠ QTR)
2(Angle TQR) = 180° - 150°
2(Angle TQR) = 30°
Angle TQR = 30/2
Angle TQR = 15°
Hence proved.
Since PQRS is a square,
angle PSR = angle QRS... (each 90°)
Now, again in equilateral triangle SRT,
we have angle TSR = angle TRS ...(each 60°)
Angle PSR + Angle TSR = Angle QRS + Angle TRS
⇒ Angle TSP = Angle TRQ
Now in Δ TSP and Δ TRQ, we have
TS = TR (Sides of equilateral triangle)
Angle TSP = Angle TRQ... (Proved above)
PS = QR ...(Sides of square)
Δ TSP ≡ Δ TRQ
So, PT = QT Proved.
Now in Δ TQR, we have
TR = QR ....(QR = RS = TR)
Angle TQR = Angle QTR and Angle TQR + Angle QTR + Angle TRQ = 180°
Angle TQR + Angle QTR + Angle TRS + Angle SRQ = 180°
2(Angle TQR) + 60° + 90° = 180° .....(∴ ∠ TQR = ∠ QTR)
2(Angle TQR) = 180° - 150°
2(Angle TQR) = 30°
Angle TQR = 30/2
Angle TQR = 15°
Hence proved.
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