Find the decimal between consecutive whole numbers around sqrt(31) whose square will be the nearest to the number under the square root.
A
6 inches
B
5.6 inches
Practice makes perfect
We have to find the length of one side of a square poster, which has area an area of 31 square inches. Remember the formula for the area of a square with side length s.
A=s^2We know that 31 is the area of the square, so its side length is sqrt(31). To estimate sqrt(31) to the nearest whole inch we will find consecutive perfect squares around 31. The two nearest perfect squares are 25 and 36.
Now we have to determine if sqrt(31) is closer to 5 inches or 6 inches. To do this we will compare sqrt(31) with 5.5, which is the midpoint of 5 and 6.
Since 31 is more than 30.25, also sqrt(31) is more than 5.5. Therefore, sqrt(31) inches is closer to 6 inches than to 5 inches, and the approximation of sqrt(31) inches to the nearest whole inch is 6 inches.
From Part A we know that sqrt(31) is somewhere between 5.5 and 6. To approximate it to the nearest tenth we will use decimals between 5.5 and 6. Let's calculate the square of each number and compare them with 31.
Approximation
Square of Approximation
Comparison
5.5
5.5 * 5.5 = 30.25
Approximation is too low
5.6
5.6 * 5.6 = 31.36
Approximation is too high
We know that sqrt(31) is somewhere between 5.5 and 5.6. In order to estimate it to the nearest tenth, we need to find which square is closer. We will do this by finding the difference between 30.25 and 31, and 31 and 31.36.
30.25-0.75 ←31+0.36 →31.36
Because 31 is closer to 31.36, we know that sqrt(31) is closer to sqrt(31.36). Therefore, the side length of the poster approximated to the nearest tenth of an inch is 5.6 inches.
