Point P inside square ABCD Find the side of the square if the distances from P to AB and AD

Point P inside square ABCD Find the side of the square if the distances from P to AB and AD are 3 and 1, respectively, and the distance PC is √34.

Given: Square. You need to find the side of the square. According to the specification: PG = 3, PF = 1, PC = √ (34). Let x denote the side of the square ABCD.
It is clear that AG = PF = 1, therefore, PE = GB = AB – AG = x – 1. Similarly, since EB = PG = 3, then EC = BC – EB = x – 3.
The CPE triangle is a rectangular rectangle since ∠CEP = 90 °, where PC is the hypotenuse, PE and EC are the legs. According to the Pythagorean theorem, PC ^ 2 = PE ^ 2 + EC ^ 2. Therefore (√ (34)) ^ 2 = (x – 1) ^ 2 + (x – 3) ^ 2. Applying the formula for abbreviated multiplication (a – b) ^ 2 = a ^ 2 – 2 * a * b + b ^ 2 (the square of the difference), we get: x ^ 2 – 2 * x * 1 + 1 ^ 2 + x ^ 2 – 2 * x * 3 + 12 = 34 or x ^ 2 – 4 * x – 1 ^ 2 = 0.
The last quadratic equation has two roots x1 = 6 and x2 = –2 – a subsidiary root.
Answer: The side of the square is 6.