*Answer :*

*Given,*

PQRS is a square and SRT is a equilateral triangle

**To prove:** (i) PT = QT

(ii) ∠*TQR* =15 �

**Proof:** PQ = QR = RS = SP (Given) (i)

And, ∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90^{o}

And,

SRT is an equilateral triangle

SR = RT = TS (ii)

And, ∠TSR = ∠SRT = ∠RTS = 60^{o}

From (i) and (ii)

PQ = QR = SP = SR= RT = TS (iii)

∠TSP = ∠TSR + ∠RSP

= 60^{o} + 90^{o} = 150^{o}

∠TRQ = ∠TRS + ∠SRQ

= 60^{o} + 90^{o} = 150^{o}

Therefore, ∠TSR = ∠TRQ = 150^{o} (iv)

Now, in ∆TSP and ∆TRQ, we have

TS = TR (From iii)

∠TSP = ∠TRQ (From iv)

SP = RQ (From iii)

Therefore, By SAS theorem,

∆TSP ≅ ∆TRQ

PT = QT (BY c.p.c.t)

In ∆TQR

QR = TR (From iii)

Hence, ∆TQR is an isosceles triangle.

Therefore, ∠QTR = ∠TQR (Angles opposite to equal sides)

Now,

Sum of angles in a triangle is 180^{o}.

∠QTR + ∠TQR + ∠TRQ = 180^{O}

2∠TQR + 150^{O} = 180^{O} (From iv)

2∠TQR = 30^{O}

∠TQR = 15^{O}

Hence, proved

*Rate this question :*

*How useful is this solution?We strive to provide quality solutions. Please rate us to serve you better.*

*Related VideosProperties of Triangles54 minsRevision on Angle sum Property of Triangles44 minsRD Sharma | Problems on Triangles41 minsTriangles - Important Questions39 minsR.D. Sharma | Triangles33 minsNCERT | Most Important Questions of Triangles50 minsNCERT | Solve Qs. on SAS Congruency Criteria46 minsQuiz | Imp. Qs. on SSS Congruency Criterion45 minsQuiz | Learn Triangles through Questions43 minsGenius Quiz | Know About Triangles38 mins*

*Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic ExpertsDedicated counsellor for each student24X7 Doubt ResolutionDaily Report CardDetailed Performance Evaluationview all courses*

*RELATED QUESTIONS :*

Prove that the anRS Aggarwal & V Aggarwal - Mathematics

If the sides of aRD Sharma - Mathematics

D is any point onNCERT Mathematics Exemplar

In a triangle *RD Sharma - Mathematics*

In Δ *ABC**RD Sharma - Mathematics*

In Δ *PQR**RD Sharma - Mathematics*

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

Prove that the peRD Sharma - Mathematics

In Fig. 10.25, <iRD Sharma - Mathematics

In a <span lang="RD Sharma - Mathematics