Square and square roots part-4

Solutions

1. Question 1: Calculate \(\sqrt{208+\sqrt{2304}}\)

   Solution: First, calculate the square root of 2304 and then add it to 208 before taking the square root of the sum.

   \[ = \sqrt{208+\sqrt{2304}} \] \[ = \sqrt{208+48} \] \[ = \sqrt{256} \] \[ = 16 \]

2. Question 2: Calculate \(\sqrt{0.0016}\)

   Solution: The square root of 0.0016 is:

   \[ = \sqrt{0.0016} \] \[ = \sqrt{16 \times 10^{-4}} \] \[ = \sqrt{16} \times \sqrt{10^{-4}} \] \[ = 4 \times 10^{-2} \] \[ = 0.04 \]

3. Question 3: Given that \(\sqrt{1521}=39\), find the value of \(\sqrt{0.1521}+\sqrt{15.21}\).

   Solution: This seems to be an error, because \(\sqrt{1521}\) should be equal to 39. Assuming the correct value is \(\sqrt{1521} = 39\), let's calculate:

   \[ = \sqrt{0.1521} + \sqrt{15.21} \] \[ = \sqrt{\frac{1521}{10000}} + \sqrt{\frac{1521}{100}} \] \[ = \frac{39}{100} + \frac{39}{10} \] \[ = 0.39 + 3.9 \] \[ = 4.29 \]

4. Question 4: 1 + 3 + 5 + 7 + \(\cdots\) up to n terms is equal to?

   Solution: This is the sum of the first n odd numbers. The sum of the first n odd numbers is equal to \(n^2\). Therefore,

   \[ 1 + 3 + 5 + 7 + \cdots = n^2 \]