Question

to one decimal place, \\(\sqrt{97}\\) must lie between \\(\square\\) and \\(\square\\)

Explanation:

Step1: Find the perfect squares around 97

We know that $9^2 = 81$ and $10^2 = 100$. But we need to check the squares of numbers with one decimal place. Let's consider $9.8^2$ and $9.9^2$.
First, calculate $9.8^2$: $9.8\times9.8 = 96.04$.
Then, calculate $9.9^2$: $9.9\times9.9 = 98.01$.

Step2: Compare with 97

Since $9.8^2 = 96.04 < 97$ and $9.9^2 = 98.01 > 97$, so $\sqrt{97}$ lies between $9.8$ and $9.9$.

Answer:

9.8 and 9.9