Question

in the figure if traingle PTR is congruent to traingle QTS, then show that angle TQS= angle QRT+ angle QSR.​

Answer 1

Answer:

this is the answer

Step-by-step explanation:

Given that:

QP=8cm,PR=6cm and SR=3cm

(I) In △PQR and △SPR

∠PRQ=∠SRP (Common angle)

∠QPR=∠PSR (Given that)

∠PQR=∠PSR (Properties of triangle )

∴△PQR∼△SPR (By AAA)

(II)

SP

PQ

=

PR

QR

=

SR

PR

(Properties of similar triangles)

⇒

SP

8cm

=

3cm

6cm

⇒SP=4cm and

⇒

6cm

QR

=

3cm

6cm

⇒QR=12cm

(III)

ar(△SPR)

ar(△PQR)

=

SP

2

PQ

2

=

4

2

8

2

=4

Answer 2

Since,

∆PTR ≅ ∆QTS,

By CPCT

PT = QT

TR = TS

PR = QS

∠PTR = ∠QTS

∠TRP = ∠TSQ

∠RPT = ∠SQT

Let,

a = ∠TRQ

b = ∠QSR

x = ∠PTQ

Thus,

In ∆TPQ

∠TPQ = ∠TQP = 90 - $\frac{x}{2}$                     [Angle Sum and Isoceles Triangle]

Similarly,

In ∆TRS,

∠TRS = ∠TSR = 90 - $\frac{x}{2}$                     [Angle Sum and Isoceles Triangle]

Now,

90 - $\frac{x}{2}$ + 90 - [Angle sum in quadrilateral]

a = 90 - $\frac{x}{2}$ - y

Also,

90 - $\frac{x}{2}$ - y + b = 90 - [Angle S]

y = b

∠TQS = 90 - $\frac{x}{2}$

= 90 - $\frac{x}{2}$ - y + y        [Adding and subtracting 'y']

= ∠QRT + ∠QSR              [∠QRT = a, ∠QSR = b]

Therefore Proved

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