The Square Root of 100 is 10
The square root of 100 is 10. It is a perfect square: $10 \times 10 = 100$, so $\sqrt{100} = 10$ — no decimal, no rounding.
It is also the round-number anchor of the square-root table from $1$ to $100$: $\sqrt{100}$ closes the run of perfect squares that starts at $\sqrt{1} = 1$.
Quick Answer:
Result: $\sqrt{100} = 10$
Notation: radical form $\sqrt{100}$; exponent form $100^{1/2}$.
Method shown: prime factorisation, with cross-checks by long division and repeated subtraction.
Rational or irrational: rational — $10 = \tfrac{10}{1}$.
Exact form: $10$ (an integer; the radical fully resolves).
Quick Reference Table — Square Roots of Nearby Perfect Squares
$n$ | $\sqrt{n}$ | Perfect square? |
|---|---|---|
$49$ | $7$ | yes |
$64$ | $8$ | yes |
$81$ | $9$ | yes |
$100$ | $10$ | yes |
$121$ | $11$ | yes |
$144$ | $12$ | yes |
$169$ | $13$ | yes |
$196$ | $14$ | yes |
$225$ | $15$ | yes |
$256$ | $16$ | yes |
These are the perfect squares bracketing $100$, with whole-number roots on either side — useful for estimating any nearby non-perfect square at a glance.
Where The Square Root of 100 Appears
The $\sqrt{100} = 10$ result is everywhere the decimal system is, because $100 = 10^2$ is the base of our place value — hundreds, percentages, and centi- units all rest on it. Geometrically, a square of area $100$ square units has a side of $\sqrt{100} = 10$ units. It also turns up in the Pythagorean theorem: a $6$–$8$–$10$ right triangle has a hypotenuse of $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$, which is just the $3$–$4$–$5$ triangle scaled by two.
What "square root of 100" Means
A square root of a number $n$ is a value $x$ for which $x^2 = n$. For $100$, that value is $10$, since $10^2 = 100$.
The radical symbol $\sqrt{;}$ returns the principal (non-negative) root, so $\sqrt{100} = 10$. The equation $x^2 = 100$ has two solutions, $10$ and $-10$, but $\sqrt{100}$ alone names only the positive one.
How To Find The Square Root of 100
Method 1 — Prime factorisation
Factor $100$ into primes, pair them, and take one from each pair.
$$100 = 2 \times 2 \times 5 \times 5 = (2 \times 2)(5 \times 5)$$
One $2$ and one $5$ leave their pairs: $\sqrt{100} = 2 \times 5 = 10$.
Final answer: $\sqrt{100} = 10$.
Method 2 — Long division
Pair the digits from the right: $\overline{1},\overline{00}$. The largest integer whose square is at most $1$ is $1$, so the first quotient digit is $1$, with remainder $0$. Bring down $00$ to get a dividend of $0$; double the quotient ($1 \to 2$) and find $d$ with $(20 + d)\cdot d \le 0$, which gives $d = 0$. The quotient is $10$ with remainder $0$, so $\sqrt{100} = 10$ exactly.
Final answer: $\sqrt{100} = 10$.
Method 3 — Repeated subtraction of odd numbers
Subtract successive odd numbers and count the steps to reach $0$: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19$.
That is $10$ odd numbers, and $1 + 3 + \dots + 19 = 100$, so $\sqrt{100} = 10$.
Examples of Square Root of 100
Example 1
Evaluate $\sqrt{100}$ directly.
$10^2 = 100$, so $\sqrt{100} = 10$. A whole number — no estimation needed.
Example 2
A student is asked to simplify $\sqrt{100} + \sqrt{100}$.
Wrong attempt. Reading it as "the root of $100 + 100$," the student writes $\sqrt{200} \approx 14.14$.
Correct. The two radicals are separate terms: $\sqrt{100} + \sqrt{100} = 10 + 10 = 20$. Square roots are evaluated before adding — the expression is $10 + 10$, not $\sqrt{200}$.
Example 3
Simplify $\sqrt{100x^2}$ for $x > 0$.
$\sqrt{100x^2} = \sqrt{100}\cdot\sqrt{x^2} = 10x$. The root distributes over a product.
Example 4
Find the side of a square whose area is $100\ \text{cm}^2$.
side $= \sqrt{100} = 10\ \text{cm}$. Area $=$ side$^2$, so the side is the square root of the area.
Example 5
A right triangle has legs $6$ and $8$. Find the hypotenuse.
$c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. The $6$–$8$–$10$ triangle.
Common Mistakes With the Square Root of 100
Mistake 1: Confusing √100 with 100 ÷ 2
Where it slips in: A learner reaches for division instead of asking what value squares to $100$.
Don't do this: $\sqrt{100} = 50$.
The correct way: $\sqrt{100}$ asks "what number times itself gives $100$?" — that's $10$, since $10 \times 10 = 100$. Halving gives $50$, which is unrelated.
Mistake 2: Writing √100 as ±10
Where it slips in: Confusing the radical with the solutions of $x^2 = 100$.
Don't do this: $\sqrt{100} = \pm 10$.
The correct way: $\sqrt{100} = 10$. The radical returns only the principal root; the $\pm 10$ belongs to solving the equation $x^2 = 100$, not to the symbol $\sqrt{100}$.
Mistake 3: Splitting the root over a sum
Where it slips in: Treating $\sqrt{36 + 64}$ as $\sqrt{36} + \sqrt{64}$.
Don't do this: $\sqrt{36 + 64} = 6 + 8 = 14$.
The correct way: $\sqrt{36 + 64} = \sqrt{100} = 10$. Square roots do not distribute over addition — add first, then take the root.
Conclusion
The square root of 100 is $10$, an exact whole number, because $100 = 10^2$ is a perfect square.
Prime factorisation gives it directly: $100 = 2^2 \times 5^2$, so $\sqrt{100} = 2 \times 5 = 10$.
$\sqrt{100}$ names only the principal root, $10$; the equation $x^2 = 100$ is what yields $\pm 10$.
Because $10$ is an integer, $\sqrt{100}$ is rational.
$\sqrt{100} = 10$ anchors the $6$–$8$–$10$ triangle and the base-$10$ place-value system.
A Practical Next Step
List $\sqrt{81}$, $\sqrt{121}$, and $\sqrt{144}$ from memory, then verify by squaring.
Solve $x^2 = 100$ and give both solutions — then explain why $\sqrt{100}$ returns only one.
Use $\sqrt{100} = 10$ to find the hypotenuse of a $6$–$8$ legged triangle.
Want a Bhanzu trainer to walk through more square-root problems? Book a free demo class — online globally.
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