[ Rectangles and Squares. Properties and Characteristics ]

In a plane, there is a square with vertices $A, B, C, D$ in sequence and a point $O$ outside the square. It is known that $A O=O B=5$ and $O D=\sqrt{13}$. Find the area of the square.
Compose an equation with respect to the side of the square. One of the solutions of this equation contradicts the condition of the problem.

## Solution

Let $x$ be the side of the square. Let $P$ be the projection of point $O$ onto the line $A D$. From the condition of the problem, it follows that points $O$ and $A$ lie on opposite sides of the line $C D$.

Since point $O$ is equidistant from points $A$ and $B$, point $O$ lies on the perpendicular bisector of segments $A B$ and $C D$. Therefore, $O P = C D / 2 = x / 2$.

By the Pythagorean theorem in right triangles $A R O$ and $D P O$:

$$
A P = \sqrt{A O^{2} - O P^{2}} = \sqrt{25 - x^{2} / 4} ; D P = \sqrt{O D^{2} - O P^{2}} = \sqrt{13 - x^{2} / 4}
$$

Since $A D + D P = A P$, then $x + \sqrt{13 - x^{2} / 4} = \sqrt{25 - x^{2} / 4}$.

From the obtained equation, we find that $x^{2} = 2$ or $x^{2} = 36$. In the second case, $A D = 6$, which is impossible, since

$$
A P = A D + D P = 6 + D P > 5 = A O
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b0b755a355dbd05a96cdg-31.jpg?height=388&width=874&top_left_y=2290&top_left_x=595)

## Answer

2. 

Let $A, B, C, D$ be the consecutive vertices of the square, and point $O$ is located inside the square. It is known that $O C = O D = \sqrt{10}$ and $O B = \sqrt{26}$. Find the area of the square.

## Solution

See the given problem.

## Answer

32.