Square and square roots-2

Math Problems

1. Amit walks 16m south from his friend's house. While returning, he walks diagonally from his friend's house to reach back to his own house. What distance did he walk while returning?

   Solution:

   As Amit walks in a straight line to his friend's house and then diagonally back, it forms a right-angled triangle. We can use the Pythagorean theorem to find the length of the hypotenuse (the diagonal path), where the distance walked to the friend's house is one of the sides, and is equal to 16m (the other side is assumed to be of the same length, considering he reached the same point).

   Using Pythagorean theorem \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse:

   \(c = \sqrt{16^2 + 16^2} = \sqrt{512} = 16\sqrt{2} \approx 22.63m\)

2. Find the square root of the following:

   1. \( \sqrt{16169/441} \)
   2. \( \sqrt{5.774409} \)
   3. \( \sqrt{841/1521} \)

   Solution:

   1. \( \sqrt{16169/441} = \sqrt{36.66} \approx 6.05 \)
   2. \( \sqrt{5.774409} \approx 2.4 \)
   3. \( \sqrt{841/1521} = \sqrt{0.55} \approx 0.74 \)

3. \( \sqrt{208 + \sqrt{2304}} \)

   Solution:

   \( \sqrt{208 + \sqrt{2304}} = \sqrt{208 + 48} = \sqrt{256} = 16 \)

4. \( \sqrt{0.0016} = ? \)

   Solution:

   \( \sqrt{0.0016} = 0.04 \)

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

   Solution:

   \( \sqrt{0.1521} + \sqrt{15.21} = 0.39 + 3.9 = 4.29 \)

6. \( 12.1 + 3 + 5 + 7 + \cdots \) up to n terms is equal to ?

   Solution:

   This is an arithmetic series where the first term \( a = 3 \), the common difference \( d = 2 \), and the number of terms \( n \) is given. The sum of an arithmetic series can be calculated by the formula \( S = \frac{n}{2}[2a + (n-1)d] \). However, this series has an additional term (12.1) that is not part of the arithmetic progression. So, the sum up to \( n \) terms will be:

   \( 12.1 + \frac{n}{2}[2*3 + (n-1)*2] = 12.1 + n[3 + n - 1] = 12.1 + 3n + n^2 - n = 12.1 + 2n + n^2 \).