Question

6. In the Fig., ABCD is a square and angle PQR = 90°. If
PB = QC = DR, prove that:
(ii) PQ = QR
(1) QB = RC
(iii) angleQPR = 45°​

Answer 1

Since PBCR becomes rectangle 

∴ PB=RC=x 

⇒AP=x

If QC=RC=x ∴∠CRQ=∠RCQ=45^{0}450

Since ∠PRC=90^{0}900

∴∠PRQ=45^{0}450

and sum of angles a triangle is 180^{0}1800

∴∠PQR=90^{0}900  and ∠PRQ=45^{0}450

∴∠QPR=180^{0}1800 -(90^{0}900 +45^{0}450 )=45^{0}450