Given a square ABCD and a point O outside the square. It is known that ОА = ОВ = 5, ОD = √13. Find the area of the square.

| September 4, 2021 | education

Let point О be located outside the square and ОА = ОВ = 5, ОD = √ (13). At the request of the task, we calculate the area of ​​the square ABCD. As you know, the area of ​​the square S is equal to the square of the side, that is, S = AB².
First of all, note that triangle AOB is an isosceles triangle, since OA = OB = 5. Then its height OF is also the median. Therefore, AF = FB = AB / 2.
The point of intersection of the segment OF with the side of the square CD is denoted by E. Since the distance between the points is preserved during the parallel transfer, moving point A to D, and point F to E, we obtain DE = AF = AB / 2. We also state the fact that EF = AD = AB.
For brevity, we introduce the notation AF = x and OE = y. Then, ОF = ЕF + OE = 2 * x + y and DE = AF = x. We apply twice the Pythagorean theorem to right-angled triangles: ΔOED and ΔOFA. We have: ОD² = DE² + OE² and ОА² = AF² + OF² or, in the new notation and data, (√ (13)) ² = x² + y² and 5² = x² + (2 * x + y) ² or x² + y² = 13 (first equation) and 5 * x² + 4 * x * y + y² = 25 (second equation). Let us solve the resulting system of equations for the unknowns x and y (taking into account the physical meaning).
Using the first equation from the second we get 13 + 4 * x² + 4 * x * y = 25 or x² + x * y = 3, whence x * y = 3 – x² and x * (x + y) = 3. Square both sides of the last equality: x² * (x + y) ² = 9 or x² * (x² + y² + 2 * x * y) = 9, whence x² * (13 + 2 * (3 – x²)) = 9. Simplify : 2 * x⁴ – 19 * x² + 9. This biquadratic equation has four roots: x1 = -3; x2 = 3; x3 = -√ (2) / 2 and x4 = √ (2) / 2. We discard negative roots immediately as side ones.
Let x = 3. Then the equality x * (x + y) = 3 will get the form 3 * (3 + y) = 3 or 3 + y = 1, whence y = 1 – 3 = -2, which is impossible. Let us examine the last root x = √ (2) / 2. In this case, the equality x * (x + y) = 3 takes the form (√ (2) / 2) * ((√ (2) / 2) + y) = 3 or √ (2) / 2 + y = 3√ (2), whence y = 5√ (2) / 2. The result obtained allows us to assert that the side of the square ABCD is equal to AB = 2 * AF = 2 * x = 2 * ( √ (2) / 2) = √ (2). Therefore, its area S = AB² = (√ (2)) ² = 2.
Answer: 2.



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