**Telangana**** SCERT Solution Class IX (9) Math Chapter 12 Circles Exercise 12.2**

**(1) In the figure, if AB =CD and ****∠****AOB = 90 ^{o} find **

**∠**

**COD**

**Solution:**

Given, AB = CD

∠AOB = 90^{o}

AB & CD are two equal chords of the circle

∴ ∠COD = ∠AOB = 90^{o} [∵angle subtended by two equal chords at the centre is equal]

**(2.) In the figure, PQ = RS and ****∠****ΟRS = 48 ^{o} . Find **

**∠**

**OPQ and**

**∠**

**ROS**

**Solution:**

Given, PQ = RS

∠ORS= 48^{o}

In △POQ & △ROS

(i) ∠ROS = ∠POQ [∵angle subtended by equal chords of a circle at the centre is equal]

(ii) RO = OP [radius of circle]

(iii) OS = OQ [radius of circle]

∴ △POQ ≅ △ROS by SAS

∴ ∠OPQ = ∠ORS = 48^{o} [corresponding angles of congruent triangle are equal]

∠RSO = ∠PQO [corresponding angles of congruent △S are equal]

OR = OS [∵radius of circle]

∴In △ROS, △ORS = ∠OSR = 48^{o} [opposite angles of equal sides are equal]

∴ ∠ORS + ∠OSR + ∠ROS = 180^{o} [sum of angles of △ is 180^{o}]

Or, 48 + 48 + ROS = 180^{o}

Or, <ROS = 180^{o} – 96^{o}

Or, <ROS = 84^{o}

**(3) In the figure PR and QS are two diameters. Is PQ = RS?**

**Solution:**

Given, PR and QS are two diameter of the circle

∴ O is the centre of circle since the intersection of two diameter of a circle is the centre.

∴ In △POQ & △ROS

(i) PO = OR [radius of circle]

(ii) OQ = SO [radius of circle]

(iii) ∠POQ = ∠ROS [vertically opposite angles]

∴ △POQ ≅ △ROS by SAS

∴PQ = RS [∵corresponding sides of congruent △S are equal]