Question:
In a city, there is a circular park. There are four points of entry into the park, namely P, Q, R, and S. Three paths were constructed which connected the points PQ, RS, and PS. The length of the path PQ is 10 units, and the length of the path RS is 7 units. Later, the municipal corporation extended the paths PQ and RS past Q and R respectively, and they meet at a point T on the main road outside the park. The path from Q to T measures 8 units, and it was found that the angle PTS is 60°. Find the area (in square units) enclosed by the paths PT, TS, and PS.
In a city, there is a circular park. There are four points of entry into the park, namely P, Q, R, and S. Three paths were constructed which connected the points PQ, RS, and PS. The length of the path PQ is 10 units, and the length of the path RS is 7 units. Later, the municipal corporation extended the paths PQ and RS past Q and R respectively, and they meet at a point T on the main road outside the park. The path from Q to T measures 8 units, and it was found that the angle PTS is 60°. Find the area (in square units) enclosed by the paths PT, TS, and PS.
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In circle geometry, when secants extend from an external point, use the Secant-Secant Theorem: $(\text{external segment}) \times (\text{whole secant})$ for both lines. This simplifies finding unknown lengths.
Updated On: Aug 23, 2025
- $36\sqrt{3}$
- $54\sqrt{3}$
- $72\sqrt{3}$
- $90\sqrt{3}$
- None of the above
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The Correct Option is C
Solution and Explanation
Step 1: Information given.
- PQ = 10 units
- RS = 7 units
- QT = 8 units
- $\angle PTS = 60^\circ$ We are required to find the area of $\triangle PTS$.
Step 2: Use secant-secant theorem.
From an external point T, if two secants intersect the circle, then: \[ TQ \times TP = TR \times TS \] where $TP, TQ$ are lengths along one secant, and $TR, TS$ along the other. Here: \[ TQ \times (TQ + PQ) = TR \times (TR + RS) \] Substituting values: \[ 8 \times (8 + 10) = TR \times (TR + 7) \] \[ 8 \times 18 = TR^2 + 7TR \] \[ 144 = TR^2 + 7TR \]
Step 3: Solve quadratic for TR.
\[ TR^2 + 7TR - 144 = 0 \] Factorizing: \[ TR^2 + 7TR - 144 = (TR+16)(TR-9) = 0 \] So, $TR = 9$ (positive root).
Step 4: Find TS.
\[ TS = TR + RS = 9 + 7 = 16 \]
Step 5: Find PT.
\[ PT = PQ + QT = 10 + 8 = 18 \]
Step 6: Area of triangle PTS.
We now know: - $PT = 18$, - $TS = 16$, - $\angle PTS = 60^\circ$. Area formula for a triangle: \[ \text{Area} = \tfrac{1}{2} \times PT \times TS \times \sin \angle PTS \] \[ = \tfrac{1}{2} \times 18 \times 16 \times \sin 60^\circ \] \[ = 144 \times \tfrac{\sqrt{3}}{2} = 72\sqrt{3} \]
Final Answer:
\[ \boxed{72\sqrt{3}} \]
- PQ = 10 units
- RS = 7 units
- QT = 8 units
- $\angle PTS = 60^\circ$ We are required to find the area of $\triangle PTS$.
Step 2: Use secant-secant theorem.
From an external point T, if two secants intersect the circle, then: \[ TQ \times TP = TR \times TS \] where $TP, TQ$ are lengths along one secant, and $TR, TS$ along the other. Here: \[ TQ \times (TQ + PQ) = TR \times (TR + RS) \] Substituting values: \[ 8 \times (8 + 10) = TR \times (TR + 7) \] \[ 8 \times 18 = TR^2 + 7TR \] \[ 144 = TR^2 + 7TR \]
Step 3: Solve quadratic for TR.
\[ TR^2 + 7TR - 144 = 0 \] Factorizing: \[ TR^2 + 7TR - 144 = (TR+16)(TR-9) = 0 \] So, $TR = 9$ (positive root).
Step 4: Find TS.
\[ TS = TR + RS = 9 + 7 = 16 \]
Step 5: Find PT.
\[ PT = PQ + QT = 10 + 8 = 18 \]
Step 6: Area of triangle PTS.
We now know: - $PT = 18$, - $TS = 16$, - $\angle PTS = 60^\circ$. Area formula for a triangle: \[ \text{Area} = \tfrac{1}{2} \times PT \times TS \times \sin \angle PTS \] \[ = \tfrac{1}{2} \times 18 \times 16 \times \sin 60^\circ \] \[ = 144 \times \tfrac{\sqrt{3}}{2} = 72\sqrt{3} \]
Final Answer:
\[ \boxed{72\sqrt{3}} \]
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