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 lines joining the poles AB, BC, CD and DA touch 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 $\angle \text{QOR}$. (tangent, circle, quadrilateral, angle, radius)

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): Properties of tangents from an external point

Tangents drawn from an external point to a circle are equal in length. For the quadrilateral ABCD, the sides AB, BC, CD, DA are tangents to the circle at points P, Q, R, S respectively. Thus, we have four pairs of equal tangent segments.

Step 3: Part (ii): Identifying the quadrilateral

Given that $\text{AB} = \text{AD}$. We can substitute the tangent segment equalities into this equation. Since $\text{AP} = \text{AS}$ and $\text{PB} = \text{BQ}$, and $\text{SD} = \text{DR}$, the condition $\text{AB} = \text{AD}$ implies that $\text{AP} + \text{BQ} = \text{AP} + \text{DR}$. This simplifies to $\text{BQ} = \text{DR}$. A quadrilateral circumscribing a circle where adjacent sides are equal (like AB=AD) is a kite.

Step 4: Part (iii) (a): Calculating AP

From the property that tangents from an external point are equal, we know that $\text{AP} = \text{AS}$ and $\text{DR} = \text{DS}$. The side $\text{AD}$ is composed of segments $\text{AS}$ and $\text{SD}$. Therefore, $\text{AD} = \text{AP} + \text{DR}$. We are given $\text{AD} = 11 \text{ cm}$ and $\text{DR} = 7 \text{ cm}$. Substituting these values, we can find $\text{AP}$.

Step 5: Part (iii) (a): Final calculation for AP

Substituting the given values into the equation $\text{AD} = \text{AP} + \text{DR}$, we get $11 = \text{AP} + 7$. Solving for $\text{AP}$ gives us $4 \text{ cm}$.

Step 6: Part (iii) (b): Angle at the center

Consider the quadrilateral OQCR. We know that $\angle \text{OQC} = 90^\circ$ and $\angle \text{ORC} = 90^\circ$ because the radius is perpendicular to the tangent at the point of contact. The sum of angles in a quadrilateral is $360^\circ$. So, $\angle \text{QOR} + \angle \text{OQC} + \angle \text{QCR} + \angle \text{ORC} = 360^\circ$. This simplifies to $\angle \text{QOR} + 90^\circ + \angle \text{QCR} + 90^\circ = 360^\circ$, which means $\angle \text{QOR} + \angle \text{QCR} = 180^\circ$. Given $\angle \text{QCR} = 60^\circ$.

Step 7: Part (iii) (b): Final calculation for QOR

Using the relationship $\angle \text{QOR} + \angle \text{QCR} = 180^\circ$ and substituting $\angle \text{QCR} = 60^\circ$, we find $\angle \text{QOR} = 180^\circ - 60^\circ = 120^\circ$.

More from this chapter

• 17. There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the b
• In figure 10.3 if oa is equals to 5 cm ab = 8 cm
• A d is a diameter of a circle and a b is a chord if a d is equals to 34 CM ab is equals to 30 cm the distance of a b from the centre of the circle is dash
• Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
• A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find the length of PQ.
• From an external point P, two tangents PA and PB are drawn to a circle with centre O. If angle APB = 70 degrees, find angle AOB.
• If PA and PB are tangents from an external point P to a circle with centre O such that APB = 70^°, find AOB
• Find the length of a tangent from a point Q at a distance of 25 cm from the center, given that the radius of the circle is 7 cm (or 24 cm tangent length with 25

All Circles solutions →