\(PQ = OQ = 5\) Column A: The area of region OPQ Column B: 10
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In geometry problems for quantitative comparison, if a point's location is not fixed but is only constrained to a path (like a circle), test different possible locations on that path. If the quantity you're evaluating changes, the answer is likely (D).
The relationship cannot be determined from the information given.
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The Correct Option isD
Solution and Explanation
Step 1: Understanding the Concept:
We need to determine the area of triangle OPQ and compare it with 10. The area of a triangle can be found using the formula \(\frac{1}{2} \times \text{base} \times \text{height}\). Step 2: Detailed Explanation:
Let's analyze the given information:
\begin{itemize}
\item O is the origin (0, 0).
\item Q is a point on the positive x-axis. Since OQ = 5, the coordinates of Q are (5, 0).
\item P is a point (x, y) in the first quadrant.
\item PQ = 5. This means the distance from P(x, y) to Q(5, 0) is 5.
\end{itemize}
Let's calculate the area of triangle OPQ. We can use OQ as the base.
Base = OQ = 5.
The height of the triangle is the perpendicular distance from vertex P to the line containing the base (the x-axis). This height is equal to the y-coordinate of P.
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times y\).
To find the area, we need the value of \(y\). We know that P(x, y) lies on a circle with center Q(5, 0) and radius 5, because \(PQ=5\). The equation of this circle is \((x-5)^2 + (y-0)^2 = 5^2\), or \((x-5)^2 + y^2 = 25\).
Since P can be any point on this circle (in the first quadrant), its y-coordinate can vary. Case 1: If P is very close to the x-axis, for example, \(P \approx (9.9, 0.1)\), then \(y\) is very small, and the area is close to 0. In this case, Area<10. Case 2: If P is at the highest point of the circle in the first quadrant, its coordinates would be (5, 5). Let's check if this point is valid: \(PQ = \sqrt{(5-5)^2 + (5-0)^2} = \sqrt{0^2 + 5^2} = 5\). So, P could be (5, 5).
In this case, the height \(y=5\), and the Area = \(\frac{1}{2} \times 5 \times 5 = 12.5\). Here, Area>10. Case 3: Can the area be exactly 10? Area = \(\frac{5}{2}y = 10 \implies y = 4\). If \(y=4\), we can find x from the circle equation: \((x-5)^2 + 4^2 = 25 \implies (x-5)^2 + 16 = 25 \implies (x-5)^2 = 9 \implies x-5 = \pm 3\). So \(x=8\) or \(x=2\). The point (2,4) is a valid location for P, and it gives an area of 10.
Since the area can be less than, equal to, or greater than 10, the relationship cannot be determined. Step 3: Final Answer:
The relationship cannot be determined from the information given.
