Question

4.
2
A
R
In fig 3.27, the circles with centres P
and Q touch each other at R.
A line passing through R meets the
circles at A and B respectively. Prove
that -
(1) seg AP || seg BQ,
(2) A APR – ARQB, and
(3) Find Z RQB if Z PAR = 35°
P Р
Fig. 3
B
21
heilas with​

Answer 1

Answer:

As AP and PR are radius of circle,

∴ AP = PR

Angle PAR = Angle PRA     ……. isosceles triangle property

Angle PAR + Angle PRA + Angle RPA = 180

∴ Angle PAR + Angle PAR + Angle RPA = 180

2 (Angle PAR) + Angle RPA = 180 ………1

 

Similarly, QR and QB are radius of circle

∴ QR = QB

Angle QRB = Angle RBQ

Angle QRB + Angle RBQ + Angle RQB = 180

∴ Angle QRB + Angle QRB + Angle RQB = 180

∴ 2 (Angle QRB) + Angle RQB = 180 …………2

 

As ARQ and ARQ form a straight line, Angle ARP = Angle QRB …..3

……. Opposite angles are same

 

From 1, 2, and 3

2 (Angle PAR) + Angle RPA = 2 (Angle QRB) + Angle RQB

∴ Angle PAR = Angle QRB

Hence

Angle QRB = Angle RBQ = Angle PAR = Angle ARP

 

1.       As the opposite angles of the triangle are equivalent,

The opposite sides are parallel to each other

 

2.       As all the angles of triangle APR is equivalent to all the angles of triangle RQB,

the triangle APR is similar to triangle RQB

 

3.       In triangle APR,

angle PAR = angle APR

angle ARP = 35

Angle ARP = Angle QRB

∴ Angle QRB = 35 degree

 

In triangle RQB,

Angle QBR = Angle QRB

∴ Angle QBR = 35

 

Angle QRB + Angle RBQ + Angle RQB = 180

∴ 35 + 35 + Angle RQB = 180

∴ Angle RQB = 180-70

∴ Angle RQB = 110 degree

Step-by-step explanation: