**Q)** **In a park, four poles are standing at positions A, B, C and D around the circular fountain such that the cloth joining the poles AB, BC, CD and DA touches the circular fountain at P, Q, R and S respectively as shown in the figure.**

**Based on the above information, answer the following questions :**

**(i) If O is the centre of the circular fountain, then ∠ OSA =**

**(ii) If AB = AD, then write the name of the figure ABCD.**

**(iii) (a) If DR = 7 cm and AD = 11 cm, then find the length of AP.**

**OR (iii) (b) If O is the centre of the circular fountain with ∠ QCR = 60°, then find the measure of ∠ QOR.**

A**ns: (i) Value of ∠ OSA:**

Let’s connect OS in the diagram:

We know that the radius is always perpendicular to the tangent at the point of contact,

∴ ∠ OSA = 90°

**Hence value of ∠ OSA is 90°**

**(ii) Name of Quadrilateral ABCD:**

We are given that AB = AD and since a circle is inscribed in the quadrilateral ABCD, its all sides will be equal.

Therefore, either ABCD is a square or a Rhombus.

We also know that the Square has all 4 angles as right angles, while in the diagram, we can see the one of opposite angles is obtuse and the other set of opposite angle is acute. This qualifies for quadrilateral ABCD to be a Rhombus, not a square.

**Therefore, the quadrilateral ABCD to a Rhombus.**

**(iii) (a) Length of AP: **Let’s check the diagram again:

Here, DR and DS are the tangents to the circle from an external point D,

Therefore DR = DS (tangents to a circle from a point are equal)

Given that DR = 7 cm, ∴ DS = 7 cm

Next, we are given AD = 11 cm

∵ AD = AS + DS

∴ AS = AD – DS = 11 – 7 = 4 cm

Next, we have AP and AS tangents to the circle from an external point A, therefore AP = AS

Since, we just calculated, AS = 4 cm, ∴ AP = 4 cm

**Therefore, length of AP is 4 cm.**

**(iii) (b) Value of ∠ QOR:**

Let’s connect OQ and OR:

OQCR forms a quadrilateral.

Next, as stated above, the radius is always perpendicular to the tangent at the point of

contact, ∴ ∠ ORC = 90° and ∠ OQC = 90°

Next, we know that the sum of all 4 angles in a quadrilateral is 360°

∴ ∠ OQC + ∠ QCR +∠ ORC + ∠ QOR = 360°

We calculated above values of ∠ ORC and ∠ OQC as 90°

and we are given that ∠ QCR = 60°

By substituting values of these angles, we get:

90° + 60° + 90° + ∠ QOR = 360°

∴ ∠ QOR = 360° – 240° = 120°

**Therefore, the value of ∠ QOR is 120°**

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