Case Study - 1 36. 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 $\angle \text{OSA} = \dots$ (ii) If $\text{AB} = \text{AD}$, then write the name of the figure ABCD. (iii) (a) If $\text{DR} = 7 \text{ cm}$ and $\text{AD} = 11 \text{ cm}$, then find the length of $\text{AP}$. OR (iii) (b) If O is the centre of the circular fountain with $\angle \text{QCR} = 60^\circ$, then find the measure of $\angle \text{QOR}$. (tangents, circle, quadrilateral, angle, inscribed circle)

Step 1: Part (i): Angle between radius and tangent

According to the tangent-radius theorem, the radius drawn to the point of tangency is perpendicular to the tangent. Here, OS is the radius and DA is the tangent at point S. Therefore, the angle formed between them, $\angle \text{OSA}$, is $90^\circ$.

Step 2: Part (ii): Identifying the quadrilateral

A quadrilateral is a kite if two pairs of equal-length sides are adjacent to each other. We are given that $\text{AB} = \text{AD}$. Also, from the property that tangents from an external point to a circle are equal in length, we have $\text{AP} = \text{AS}$, $\text{BP} = \text{BQ}$, $\text{CQ} = \text{CR}$, and $\text{DR} = \text{DS}$. If $\text{AB} = \text{AD}$, then the quadrilateral ABCD is a kite.

Step 3: Part (iii) (a): Finding length of AP

We know that tangents from an external point to a circle are equal in length. So, $\text{DS} = \text{DR} = 7 \text{ cm}$. The side $\text{AD}$ is composed of $\text{AS}$ and $\text{DS}$. Thus, $\text{AS} = \text{AD} - \text{DS} = 11 \text{ cm} - 7 \text{ cm} = 4 \text{ cm}$. Since $\text{AP}$ and $\text{AS}$ are tangents from point A, $\text{AP} = \text{AS}$. Therefore, $\text{AP} = 4 \text{ cm}$.

Step 4: Part (iii) (b): Finding angle QOR

In quadrilateral $\text{OQCR}$, $\angle \text{OQC}$ and $\angle \text{ORC}$ are $90^\circ$ because the radius is perpendicular to the tangent at the point of contact. The sum of angles in a quadrilateral is $360^\circ$. So, $\angle \text{QOR} + \angle \text{OQC} + \angle \text{QCR} + \angle \text{ORC} = 360^\circ$. Substituting the known values, $\angle \text{QOR} + 90^\circ + 60^\circ + 90^\circ = 360^\circ$. This simplifies to $\angle \text{QOR} + 240^\circ = 360^\circ$, which gives $\angle \text{QOR} = 120^\circ$.

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