Please solve question 12
Attachments:
Answers
Answered by
0
This is the solution
Attachments:
Answered by
1
given PS ⊥ PQ…………(1)
QR ⊥PQ............(2)
PS =QR = 4 cm……………(3)
To prove :∆PQS≅∆QPR
Proof : In ∆ PQS & ∆ PQR
PS = QR [ using (3)]
∠SPQ= ∠ RQP =90° [ using (1)&(2)]
PQ is common
∆ PQS ≅ ∆ PQR [ SAS]
QR ⊥PQ............(2)
PS =QR = 4 cm……………(3)
To prove :∆PQS≅∆QPR
Proof : In ∆ PQS & ∆ PQR
PS = QR [ using (3)]
∠SPQ= ∠ RQP =90° [ using (1)&(2)]
PQ is common
∆ PQS ≅ ∆ PQR [ SAS]
Attachments:
Similar questions