Question

13. PQRS is a square of side 6 cm each and T is a mid-point of QR. What is the radius of circle inscribed in triangle TSR ?
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Answer 1

Answer:

(i) Because PQRS is a square

∠PSR=∠QRS=90

∘

Now In △SRT

∠TSR=∠TRS=60

∘

∠PSR+∠TSR=∠QRS+∠TRS

⟹∠TSP=∠TRQ

Now in △TSP and △TRQ

TS=TR

∠TSP=∠TRQ

PS=QR

Therefore , △TSP≡△TRQ

So PT=QT

(ii) Now in △TQR,

TR=QR(RQ=SR=TR)

∠TQR=∠QTR

And ∠TQR+∠QTR+∠TRQ=180

⟹∠TQR+∠QTR+∠TRS+∠SRQ=180

⟹2(∠TQR)+60+90=180 (∠TQR=∠QTR)

2(∠TQR)=30

⟹∠TQR=15