Math, asked by ishitavats582, 1 month ago

In ΔPQR shown below, the altitudes RS and QT are equal, and PQ = PR. (Note:Figure not to scale.) a) Prove that ΔTRQ ≅ ΔSQR. b) If SR = 8 cm and QR = 10 cm, find TR. Show your steps and give valid reasons.​

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Answers

Answered by kamalkpoddar
3

Answer:

Below is your answer buddy.

Please mark as brainliest.

Step-by-step explanation:

a) Given:

  • Altitudes RS and QT are equal
  • Side PQ = PR

To proof:

  • TRQ SQR

Proof:

In ∆TRQ and ∆SQR,

  1. RS = QT (Given)
  2. ∠RSQ = ∠QTR = 90°
  3. QR is common

Therefore, ∆TRQ ≅ ∆SQR (S.A.S) (Proved).

b) Given:

  • SR = 8cm
  • QR = 10cm

To find:

  • length of TR

Solution :

We know,

RSQ = ∠QTR = 90° (adjacent angles).

Therefore, TRQ and SQR are right angled triangles.

QT = 8cm

Since, QT = SR = 8cm (CPCT)(corresponding parts of congruent triangles)

QR = 10cm

Using Pythagoras Theorem,

(QT)²+(RT)²=(QR)²

or, 8²+(RT)²=10²

or, (RT)²=100-64

or, (RT)²=36

or, RT=36

or, RT=6cm

Therefore RT is 6cm.

Answered by RvChaudharY50
0

Solution :-

In ∆TRQ and ∆SQR we have,

→ ∠QTR = ∠RSQ { Each 90° }

→ ∠TRQ = ∠SQR { since PQ = PR, angle opposite to equal sides are equal in measure . }

→ QT = RS { given }

So,

→ ∆TRQ ≅ ΔSQR { By AAS congruence rule }

then,

→ TR = SQ { By CPCT }

now, in right angled ∆QSR,

→ SQ = √[QR² - SR²] { By pythagoras theorem }

→ SQ = √(10² - 8²)

→ SQ = √(100 - 64)

→ SQ = √(36)

→ SQ = 6 cm .

therefore,

→ TR = SQ

→ TR = 6 cm (Ans.)

Learn more :-

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