Step 1: Solve the innermost square root: $\sqrt{144} = 12$.

Step 2: Add to 52: $\sqrt{52 + 12} = \sqrt{64} = 8$.

Step 3: Add to 248: $\sqrt{248 + 8} = \sqrt{256} = 16$.

Step 1: Since $93^2 = 8649$, $\sqrt{86.49} = 9.3$.

Step 2: Substitute into equation: $9.3 + \sqrt{5 + x^2} = 12.3 \implies \sqrt{5 + x^2} = 3$.

Step 3: Square both sides: $5 + x^2 = 9 \implies x^2 = 4 \implies x = 2$.

Step 1: Multiply numerator and denominator by $100,000$ to clear decimals:

Step 2: Take square root: $\sqrt{\frac{28900}{121}} = \frac{170}{11} = 15\frac{5}{11}$.

Step 1: Rationalize each fraction:

Step 2: Expand the series:

Step 3: All middle terms cancel out, leaving:

Step 1: Combine the last two terms using a common denominator:

Step 2: Add back to the expression:

Answer: (d) 3

Step 1: Calculate $x + y$:

Step 2: Calculate $x \cdot y = 1$.

Step 3: Use identity $x^2 + y^2 = (x+y)^2 - 2xy = 8^2 - 2(1) = 64 - 2 = 62$.

Step 1: Let $x = \sqrt{6+x}$.

Step 2: Square both sides: $x^2 = 6 + x \implies x^2 - x - 6 = 0$.

Step 3: Factorize: $(x - 3)(x + 2) = 0 \implies x = 3$ (since $x > 0$).

(Quick Trick: Factorize 6 into consecutive numbers $2 \times 3$. For addition, the answer is the larger factor = 3).

Step 1: Find prime factorization of $4320$:

Step 2: Group exponents in multiples of 3:

• For $2^5$, we need one more $2$ to make $2^6$.

• For $3^3$, it's already a perfect cube.

• For $5^1$, we need two more $5$'s ($5^2 = 25$) to make $5^3$.

Step 3: Required multiplier = $2 \times 5^2 = 2 \times 25 = 50$.

Step 1: Observe options: $200^2 = 40000$, $300^2 = 90000$. So answer is in 200s.

Step 2: Units digit ends in 4, so root must end in 2 or 8.

Step 3: $230^2 = 52900$. So $\sqrt{53824}$ must be slightly above 230 $\implies 232$

Step 1: Test options: $200^2 = 40000$.

Step 2: $203^2 = (200 + 3)^2 = 40000 + 1200 + 9 = 41209$.