Question

Approximate the number to the nearest (a) integer and (b) tenth: $\sqrt{\frac{27}{4}}$.

Answer 1

Given number: $\sqrt{\frac{27}{4}}$.

Since, $\sqrt{\frac{27}{4}}=\sqrt{6.75}$

We know that

$\sqrt{4}=2$ and $\sqrt{9}=3$

So, $\sqrt{\frac{27}{4}}$ will lie in between $2$ and $3$.

Now, $\sqrt{\frac{27}{4}}=2.5980...$

(a) Approximate the numbers $\sqrt{\frac{27}{4}}$ to the nearest integer.

Let $\sqrt{\frac{27}{4}}=2.59$

The value of $\sqrt{\frac{27}{4}}$ lies between $2$ and $3$ but is more closer to the integer $3$.

Approximate value of $\sqrt{\frac{27}{4}}$ to the nearest integer is $3$.

(b) Approximate value of $\sqrt{\frac{27}{4}}$ to the nearest tenth:

Round the number to the nearest tenth by removing the digits after one digit of decimal point.

Approximate value of $\sqrt{\frac{27}{4}}$ to the nearest tenth is $2.5$.

Hence, approximate value of $\sqrt{\frac{27}{4}}$ to the nearest integer is $3$.

and approximate value of $\sqrt{\frac{27}{4}}$ to the nearest tenth is $2.5$.