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

Proved that the given conditions PT = QT and  ∠TQR =15°.

Step-by-step explanation:

Given Data

PQRS is a square

SRT is an equilateral triangle

Prove that  (i) PT = QT  and (ii) ∠TQR =15°

In the given Square,

$\text{PQ} = \text{QR } = \text{ RS} = \text{SP} \ldots \ldots \ldots(1)$

$\begin{aligned}&\text { And } \angle \text { SPQ} =\angle \text { PQR}\\&= \angle \text { QRS}=\angle \text { RSP}=90^{\circ}=\text { right angle }\end{aligned}$

$\Rightarrow\angle \text{QRS}=\angle \text{RSP}=90^{\circ}=\text { right angle }$

SRT is an equilateral triangle.

$\begin{array}{l}\Rightarrow \text{SR} = \text{RT} = \text{TS} \quad \ldots(2) \\ \angle \text{TSR} =\angle \text{SRT} =\angle \text{RTS} =60^{\circ}\end{array}$

From the equations (1) and (2) , we  get

$\text{PQ} = \text{QR} = \text{SP} = \text{ SR}= \text{RT}= \text{TS} \ldots \ldots(3)$

Also we can get,

$\begin{aligned}\angle \text{TSR} &=\angle \text{TSR} +\angle \text{RSP} \\&=60^{\circ}+90^{\circ}+150^{\circ} \\\angle \text{RQ} &=\angle \text{TRS} +\angle \text{SRQ} \\&=60^{\circ}+90^{\circ}+150^{\circ}\end{aligned}$

$\Rightarrow \angle \text{TSR} =\angle \text{TRQ} =150^{\circ} \ldots(4)$

Now in triangle TSR and Triangle TRQ,

TS = TR                      [ From the equation (3) ]

∠TSP = ∠TRQ                 [ From the equation (4) ]

SP = RQ                      [ From the equation (3) ]

By SAS congruence criterion, we have

ΔTSP ≅ ΔTRQ

We know that the corresponding parts of congruent triangles are equal. Then,   PT = QT

(ii) Consider   ΔTQR ,

QR=TR                  [From(3)]

ΔTQR is a isosceles triangle

∠QTR= ∠TQR       [angles opposite to equal sides]

Sum of angles in a triangle is equal to 180°. Then,

$\Rightarrow \angle \text{QTR}+\angle \text{TQR} +\angle \text{TRQ}=180^{\circ}$

$\Rightarrow 2 \angle \text{TQR} +150^{\circ}=180^{\circ} \quad[\text { From the eqaution }(4)]$

$\Rightarrow 2 \angle \text{TQR} =180^{\circ}-150^{\circ}$

$\Rightarrow 2 \angle \text{TQR} =30^{\circ} \\\Rightarrow \angle \text{TQR} =15^{\circ}$

Hence the given condition has been satisfied.

To Learn More

1) In a equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes

https://brainly.in/question/37132

2) DEFG is a square and ∠BAC = 90º. Prove that : (a) ∆ AGF ~ ∆DBG (b) ∆AGF ~ ∆EFC (c) ∆DBG ~ ∆EFC (d) DE2 = BD × EC

https://brainly.in/question/759200

Answer 2

i

Answer:

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∘