Approximate 1131√

Question

, correct to one decimal place, by using a sequence of values that gets closer and closer to 31−−√
.(1 point)

Answer 1

To approximate $\sqrt{1131}$ and demonstrate a method that gets closer to $\sqrt{31}$, we can use the fact that $\sqrt{1131}$ is approximately equal to $\sqrt{31}\times \sqrt{37}$ (since $1131=31\times 37$).

First, we need to calculate $\sqrt{31}$ to a one-decimal place:

1. Estimate $\sqrt{31}$: We know that ${5.5}^2=30.25$ and ${5.6}^2=31.36$. Hence, $\sqrt{31}$ is between $5.5$ and $5.6$.

2. Refining the estimate: Let's try $5.57$: $${5.57}^2=31.0649(\text{too high})$$

   Now try $5.56$: $${5.56}^2=30.9136(\text{too low})$$

   Since $5.56<\sqrt{31}<5.57$, we can say that $\sqrt{31}\approx 5.57$ to one decimal place.

3. Finding $\sqrt{1131}$: Since $\sqrt{1131}=\sqrt{31\times 37}=\sqrt{31}\cdot \sqrt{37}$ and we have $\sqrt{31}\approx 5.57$, we will also need to estimate $\sqrt{37}$.

4. Estimate $\sqrt{37}$: Similarly, we know that $6^2=36$ and $7^2=49$. Thus, $\sqrt{37}$ is between $6$ and $7$.

   Let's try $6.1$: $${6.1}^2=37.21(\text{too high})$$ And then $6.05$: $${6.05}^2=36.6025(\text{too low})$$

   Therefore, $\sqrt{37}\approx 6.1$ to one decimal place.

5. Multiply the estimates: $$\sqrt{1131}\approx 5.57\cdot 6.1\approx 33.977$$

6. Round to one decimal place: $$\sqrt{1131}\approx 34.0$$

Thus, the approximation of $\sqrt{1131}$ correct to one decimal place is $\boxed{{\displaystyle 34.0}}$.