What is the square root of 50 rounded to the nearest hundredth?

Let's write an inequality, such as follows: x < sqrt (50) < y. Square both sides of the equation. We get x^2 < 50 < y^2. Obviously, x is between 7 and 8. Also notice, that for integers a, b, (ab) ^2/b^2, equals a^2. So let's try values, like 7.1. Using the previous fact, (7.1) ^2, equals (71) ^2/100. So, (7.1) ^2, equals 50.41. Thus, our number is between 7 and 7.1. We find, with a bit of experimentation, that the square root of 50, is 7.07.