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Question
In Fig. 10.23, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT (ii) ∠TQR = 15°
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Solution
Given that PQRS is a square and SRT is an equilateral triangle. And given to prove that
PT = QT and ∠ TQR =15 °
Now , PQRS is a square
⇒ PQ =QR=RS=SP ....................... (1)
And also, SRT is an equilateral triangle.
⇒ SR = RT=TS .............................(2)
And ∠TSR = ∠SRT= ∠RTS = 60°
From (1) and (2)
PQ=QR=SP=SR=RT=TS ...........................(3)
And also,
∠TSR=∠TSR+∠RSP= 60° +90° +150°
∠TRQ=∠TRS+∠SRQ=60°+90°+150°
⇒ ∠TSR=∠TRQ=150° ............................(4)
Now, in Δ TSR and Δ TRQ
TS=TR [from (3)]Δ
∠TSP = ∠TRQ [from (4)]
SP=RQ [from (3)]
So, by SAS ccongruence criterion we have
Δ TSR ≅ Δ TRQ
⇒ PT=QT [corresponding parts of congruent triangles are equal ]
Consider Δ TQR,
QR= TR [from (3)]
⇒ Δ TQR is a isosceles triangle
∠QTR=∠TQR [angles opposite to equal sides]
Now,
Sum of angles in a traingle is qual to 180°
⇒ ∠ QTR+∠TQR + ∠TRQ =180°
⇒ 2∠TQR+150°=180 [from (4)]
⇒ 2∠TQR = 180° -150°
⇒ 2 ∠TQR = 30° ∠TQR=15 °
∴ Hence Proved
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