can you explain the question through video---**Image Description:**
* **Type:** Geometric figure.
* **Main Elements:**
* A square labeled with vertices A, B, D, E in a counter-clockwise direction starting from the bottom left (A).
* Inside the square, a point P is labeled.
* Segments PA, PB, and PC are drawn, where C is a point on the top side ED.
* The segment PC appears to be perpendicular to the side ED (the top side of the square).
* **Labels and Annotations:** Vertices labeled A, B, D, E. Point P labeled. Point C on side ED.
**Question 365:**
* **Question Stem:** 365. A square with side 6 inches is shown. If P is a point such that the segments PA, PB, and PC are equal in length, and segment PC is perpendicular to segment ED, what is the area in square inches of triangle APB? (MATHCOUNTS 1991)
* **Options:** None provided.
* **Other Relevant Text:** (MATHCOUNTS 1991)
**Question 366:**
* **Question Stem:** 366. Show that the segment from a point P, outside circle O to the center of the circle,
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We have a square ABDE with side length 6 inches. Point P is located inside the square such that segments PA, PB, and PC are all equal in length. Additionally, segment PC is perpendicular to the top side ED of the square. Our goal is to find the area of triangle APB.
Let's set up a coordinate system to solve this problem systematically. We place the square on the coordinate plane with A at the origin (0,0), B at (6,0), D at (6,6), and E at (0,6). Since PC is perpendicular to the top side ED, which is horizontal, point P and point C must have the same x-coordinate. We can represent C as (c, 6) and P as (c, p) for some values c and p.
Now we set up equations based on the equal length condition. Since PA, PB, and PC are all equal to some value k, we can write: PC equals 6 minus p, which equals k. PA equals the square root of c squared plus p squared, which equals k. And PB equals the square root of c minus 6 squared plus p squared, which also equals k. These three equations will help us find the exact coordinates of point P.
Now let's solve the system of equations. From the condition that PA equals PB, we can eliminate the square roots and find that c equals 3. Substituting c equals 3 back into the equation PC equals PA, we get 6 minus p equals the square root of 9 plus p squared. Squaring both sides and simplifying, we find that p equals 9 over 4. Therefore, point P is located at coordinates (3, 9/4).
Finally, let's calculate the area of triangle APB. We can use the base times height formula. The base AB has length 6, and the height is the y-coordinate of point P, which is 9 over 4. Therefore, the area equals one half times 6 times 9 over 4, which simplifies to 3 times 9 over 4, giving us 27 over 4 square inches. This is our final answer.