In the given figure, ABCD is a square and ∠ PQR = 90°. If PB = QC =

1.	In the given figure, ABCD is a square and ∠ PQR = 90°. If PB = QC = DR, prove that(i) QB = RC,(ii) PQ = QR,(iii) ∠ QPR = 45°
Answer» (i) We know that the line segment QB can be written as QB = BC – QC Since ABCD is a square we know that BC = DC and QC = DR So we get QB = CD – DR From the figure we get QB = RC (ii) Consider △ PBQ and △ QCR It is given that PB = QC Since ABCD is a square we get ∠ PBQ = ∠ QCR = 90^o By SAS congruence criterion △ PBQ ≅ △ QCR PQ = QR (c. p. c. t) (iii) We know that PQ = QR Consider △ PQR From the figure we know that ∠ QPR and ∠ QRP are base angles of isosceles triangle ∠ QPR = ∠ QRP We know that the sum of all the angles in a triangle is 180^o ∠ QPR + ∠ QRP + ∠ PRQ = 180^o By substituting the values in the above equation ∠ QPR + ∠ QRP + 90^o = 180^o On further calculation ∠ QPR + ∠ QRP = 180^o – 90^o By subtraction ∠ QPR + ∠ QRP = 90^o We know that ∠ QPR = ∠ QRP So we get ∠ QPR + ∠ QPR = 90^o By addition 2 ∠ QPR = 90^o By division ∠ QPR = 45^o Therefore, it is proved that ∠ QPR = 45^o.

In the given figure, ABCD is a square and ∠ PQR = 90°. If PB = QC = DR, prove that(i) QB = RC,(ii) PQ = QR,(iii) ∠ QPR = 45°

Answer»

(i) We know that the line segment QB can be written as

QB = BC – QC

Since ABCD is a square we know that BC = DC and QC = DR

So we get

QB = CD – DR

From the figure we get

QB = RC

(ii) Consider △ PBQ and △ QCR

It is given that PB = QC

Since ABCD is a square we get

∠ PBQ = ∠ QCR = 90^o

By SAS congruence criterion

△ PBQ ≅ △ QCR

PQ = QR (c. p. c. t)

(iii) We know that PQ = QR

Consider △ PQR

From the figure we know that ∠ QPR and ∠ QRP are base angles of isosceles triangle

∠ QPR = ∠ QRP

We know that the sum of all the angles in a triangle is 180^o

∠ QPR + ∠ QRP + ∠ PRQ = 180^o

By substituting the values in the above equation

∠ QPR + ∠ QRP + 90^o = 180^o

On further calculation

∠ QPR + ∠ QRP = 180^o – 90^o

By subtraction

∠ QPR + ∠ QRP = 90^o

We know that ∠ QPR = ∠ QRP

So we get

∠ QPR + ∠ QPR = 90^o

By addition

2 ∠ QPR = 90^o

By division

∠ QPR = 45^o

Therefore, it is proved that ∠ QPR = 45^o.