Square Root of 1200 - How to Simplify √1200?

What Is A Square Root?

The square root of a number $n$ is the value $r$ that satisfies $r^2 = n$. For 1200, you are looking for the number that, multiplied by itself, gives 1200.

No whole number does this. $34^2 = 1156$ falls short and $35^2 = 1225$ overshoots, so $\sqrt{1200}$ lands between 34 and 35, close to 34.64. Every positive number has two square roots, one positive and one negative, but the symbol $\sqrt{1200}$ means the principal (positive) root. For a fuller treatment of the operation itself, see the definition of a square root.

Where Does √1200 Appear?

$\sqrt{1200}$ is the height of an equilateral triangle whose side length is 40, since that height equals $\frac{40\sqrt{3}}{2} = 20\sqrt{3}$. It also turns up when you scale the diagonal of a unit square: 1200 square units of area sits inside a square roughly 34.64 units on a side, just short of a clean 35. Any time a problem builds $\sqrt{3}$ into a whole-number frame, values like $20\sqrt{3}$ are what you land on.

Quick Reference Table Of √1200

How Do You Simplify √1200? (Prime Factorization)

Can you simplify the square root of 1200?

Yes, and prime factorization is the cleanest route. Break 1200 into primes, pull out every pair, and leave the rest under the radical.

$$1200 = 2^4 \times 3 \times 5^2$$

$$\sqrt{1200} = \sqrt{2^4 \times 5^2 \times 3}$$

$$\sqrt{1200} = 2^2 \times 5 \times \sqrt{3}$$

$$\sqrt{1200} = 20\sqrt{3}$$

The pairs $2^4$ and $5^2$ leave the radical as whole numbers ($2^2 = 4$ and $5$, giving $4 \times 5 = 20$). The lone factor of 3 has no partner, so it stays inside. That is why $\sqrt{1200} = 20\sqrt{3}$, and the technique generalises through simplifying radical expressions.

If you would rather spot the largest perfect-square factor directly, notice $1200 = 400 \times 3$, and $400 = 20^2$:

$$\sqrt{1200} = \sqrt{400 \times 3}$$

$$\sqrt{1200} = \sqrt{400} \times \sqrt{3}$$

$$\sqrt{1200} = 20\sqrt{3}$$

Both roads reach the same place. Estimating first, as covered in these square root tricks, tells you the answer should sit near 34.6 before you commit to a method.

How Do You Find √1200 By Long Division?

The long-division method gives the decimal $34.64\ldots$ digit by digit, without a calculator. Pair the digits outward from the decimal point and solve one place at a time.

Step 1: Group the digits: $\overline{12},\overline{00},.,\overline{00},\overline{00}$

Step 2: The largest square below 12 is $9 = 3^2$, so the first digit is 3, remainder $12 - 9 = 3$.

Step 3: Bring down 00 to make 300. Double the quotient (3) to get 6, then find $d$ with $(60 + d) \times d \le 300$. Here $64 \times 4 = 256$, so $d = 4$.

Step 4: The quotient is now 34, remainder $300 - 256 = 44$. Bring down 00 to make 4400.

Step 5: Double 34 to get 68, then find $d$ with $(680 + d) \times d \le 4400$. Here $686 \times 6 = 4116$, so $d = 6$, and the quotient reads 34.6.

Step 6: Continue once more and the next digit is 4, giving $\sqrt{1200} \approx 34.64$.

The digits keep coming forever, which is the fingerprint of an irrational number.

Is The Square Root Of 1200 Rational Or Irrational?

$\sqrt{1200}$ is irrational because 1200 is not a perfect square. A whole number has a rational square root only when it is a perfect square, and 1200 sits between $34^2$ and $35^2$.

There is a sharper test using the prime factorization. Write $1200 = 2^4 \times 3^1 \times 5^2$. A perfect square needs even exponents on every prime (this is what makes numbers like 400, 900, and 1156 land on whole roots, as the list of perfect squares shows), but the 3 carries an odd exponent of 1. That single odd power forces the root to be irrational, which is the same reason $\sqrt{3}$ itself, the leftover factor, is irrational. Wolfram MathWorld frames this cleanly in its entry on the irrational number.

Because the value never resolves to a clean fraction, keep it as $20\sqrt{3}$ in algebra and only switch to $34.641$ when a problem asks for a decimal.

Examples Of Square Root Of 1200

Example 1

Simplify $\sqrt{1200}$ to simplest radical form.

$$\sqrt{1200} = \sqrt{400 \times 3}$$

$$\sqrt{1200} = 20\sqrt{3}$$

Final answer: $20\sqrt{3}$.

Example 2

A student writes $\sqrt{1200} = \sqrt{100 \times 12} = 10\sqrt{12}$ and stops. Is this simplest form?

The first instinct is to grab the first perfect square you notice, and $100$ is easy to spot.

$$\sqrt{1200} = \sqrt{100 \times 12} = 10\sqrt{12}$$

That is correct arithmetic, but $\sqrt{12}$ still hides a perfect square, since $12 = 4 \times 3$.

$$10\sqrt{12} = 10 \times 2\sqrt{3} = 20\sqrt{3}$$

The rescue is to keep factoring until nothing inside the radical has a perfect-square factor left. Stopping at $10\sqrt{12}$ is the most common mark-losing error on this problem.

Final answer: $20\sqrt{3}$.

Example 3

Evaluate $3\sqrt{1200} + \sqrt{300}$.

$$3\sqrt{1200} = 3 \times 20\sqrt{3} = 60\sqrt{3}$$

$$\sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3}$$

$$60\sqrt{3} + 10\sqrt{3} = 70\sqrt{3}$$

Final answer: $70\sqrt{3} \approx 121.24$.

Example 4

A square has an area of 1200 square centimetres. What is the length of one side, to two decimal places?

$$\text{side} = \sqrt{1200}$$

$$\text{side} = 20\sqrt{3}$$

$$\text{side} \approx 34.64 \text{ cm}$$

Final answer: about $34.64$ cm.

Example 5

Simplify $\dfrac{\sqrt{1200}}{\sqrt{3}}$.

$$\frac{\sqrt{1200}}{\sqrt{3}} = \sqrt{\frac{1200}{3}}$$

$$\sqrt{\frac{1200}{3}} = \sqrt{400}$$

$$\sqrt{400} = 20$$

Final answer: $20$. Dividing out the shared factor of 3 turns an irrational-looking expression into a whole number.

Common Mistakes

Mistake 1: Stopping before the radical is fully simplified

Where it slips in: Pulling out the first perfect square you see, such as $\sqrt{1200} = 10\sqrt{12}$.

Don't do this: Leaving $10\sqrt{12}$ as the answer when $\sqrt{12}$ still contains the perfect square 4.

The correct way: Factor until the number under the radical has no perfect-square factors above 1, giving $20\sqrt{3}$. Learners who rush the factor-spotting almost always land on this half-finished form.

Mistake 2: Treating $\sqrt{1200}$ as if it equals $\sqrt{400} + \sqrt{3}$

Where it slips in: Splitting a product incorrectly across addition.

Don't do this: Writing $\sqrt{400 \times 3} = \sqrt{400} + \sqrt{3}$.

The correct way: The radical of a product splits across multiplication, not addition: $\sqrt{400 \times 3} = \sqrt{400} \times \sqrt{3} = 20\sqrt{3}$. The confusion between multiplying and adding under the root is one of the most persistent habits to unlearn.

Mistake 3: Rounding to 34.64 too early

Where it slips in: Multi-step problems that carry $\sqrt{1200}$ through several operations.

Don't do this: Replacing $\sqrt{1200}$ with $34.64$ at the start and multiplying that rounded value repeatedly.

The correct way: Keep the exact form $20\sqrt{3}$ until the final step. Early rounding compounds error, and a slip here is exactly the kind of mistake that sank the Mars Climate Orbiter in 1999, when mismatched rounded and exact quantities across two teams cost a $125 million spacecraft.

Conclusion

• The square root of 1200 is $20\sqrt{3}$, roughly $34.641$.

• 1200 is not a perfect square, so $\sqrt{1200}$ is irrational.

• Prime factorization ($1200 = 2^4 \times 3 \times 5^2$) and the largest-perfect-square shortcut ($1200 = 400 \times 3$) both give $20\sqrt{3}$.

• Keep the exact radical form through multi-step work and round only at the end.

To keep building radical-simplification fluency with a teacher, work through it with an algebra tutor or join structured online math classes. Prefer exam-level pacing? A high school math tutor can pair this with surds practice, or you can book a free demo class.

Read More

• Square Root of 12 - the smaller cousin that also simplifies to a multiple of $\sqrt{3}$.

• Square Root of 3 - the irrational factor left inside $20\sqrt{3}$.

• Square Root of 1000 - a neighbouring four-digit root worked the same way.

• Squares and Square Roots - the foundations behind perfect squares and roots.

• Square Root 1 to 30 - the full reference table of roots in this range.