To find the closest decimal approximation for \( \sqrt{120} \), we can start by estimating the square roots of perfect squares around 120.
The perfect squares near 120 are:
- \( 10^2 = 100 \)
- \( 11^2 = 121 \)
From this, we know that: \[ 10 < \sqrt{120} < 11 \]
Next, we can refine our estimate. Let's try calculating \( 10.5^2 \): \[ 10.5^2 = 110.25 \] Since \( 110.25 < 120 \), we know that \( \sqrt{120} > 10.5 \).
Now let's try \( 10.9^2 \): \[ 10.9^2 = 118.81 \] Since \( 118.81 < 120 \), we can conclude that \( \sqrt{120} > 10.9 \).
Next, let's try \( 10.95^2 \): \[ 10.95^2 = 120.6025 \] Since \( 120.6025 > 120 \), we know that \( \sqrt{120} < 10.95 \).
Now let's try \( 10.92^2 \): \[ 10.92^2 = 119.2064 \] Since \( 119.2064 < 120 \), we know that \( \sqrt{120} > 10.92 \).
Next, let's check \( 10.93^2 \): \[ 10.93^2 = 119.4649 \] Since \( 119.4649 < 120 \), we also know \( \sqrt{120} > 10.93 \).
Finally, let’s check \( 10.94^2 \): \[ 10.94^2 = 119.6356 \] Since \( 119.6356 < 120 \), we conclude that \( \sqrt{120} > 10.94 \).
Summarizing our findings:
- We have \( 10.9 < \sqrt{120} < 10.95 \) and \( 10.93 < \sqrt{120} < 10.94 \).
- Given \( 10.94^2 < 120 \) and \( 10.95^2 > 120 \), we can conclude that the closest decimal approximation for \( \sqrt{120} \) without exceeding 120 is approximately \( 10.94 \).
So, the closest decimal approximation for \( \sqrt{120} \) without exceeding 120 is approximately: \[ \boxed{10.94} \]
1 answer