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.
Answers
Answer:
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Step-by-step explanation:
a) Given:
- Altitudes RS and QT are equal
- Side PQ = PR
To proof:
- ∆TRQ ≅ ∆SQR
Proof:
In ∆TRQ and ∆SQR,
- RS = QT (Given)
- ∠RSQ = ∠QTR = 90°
- 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.
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.)
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