Question

In Fig. 10.23, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT (ii) ∠TQR = 15° 

 

Answer 1

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