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 ∠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.
(tangent, circle, quadrilateral, angle, radius)
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Step-by-Step Solution
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, ∠OSA, is 90∘.
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 AB=AD. We can substitute the tangent segment equalities into this equation. Since AP=AS and PB=BQ, and SD=DR, the condition AB=AD implies that AP+BQ=AP+DR. This simplifies to BQ=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 AP=AS and DR=DS. The side AD is composed of segments AS and SD. Therefore, AD=AP+DR. We are given AD=11 cm and DR=7 cm. Substituting these values, we can find AP.
Step 5: Part (iii) (a): Final calculation for AP
Substituting the given values into the equation AD=AP+DR, we get 11=AP+7. Solving for AP gives us 4 cm.
Step 6: Part (iii) (b): Angle at the center
Consider the quadrilateral OQCR. We know that ∠OQC=90∘ and ∠ORC=90∘ because the radius is perpendicular to the tangent at the point of contact. The sum of angles in a quadrilateral is 360∘. So, ∠QOR+∠OQC+∠QCR+∠ORC=360∘. This simplifies to ∠QOR+90∘+∠QCR+90∘=360∘, which means ∠QOR+∠QCR=180∘. Given ∠QCR=60∘.
Step 7: Part (iii) (b): Final calculation for QOR
Using the relationship ∠QOR+∠QCR=180∘ and substituting ∠QCR=60∘, we find ∠QOR=180∘−60∘=120∘.