Solve (frac{sqrt{12}}{2})^2+6^2

$$( \frac { \sqrt { 12 } } { 2 } ) ^ { 2 } + 6 ^ { 2 }$$

Evaluate

$$39$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(\frac{2\sqrt{3}}{2}\right)^{2}+6^{2}$$

Cancel out 2 and 2.

$$\left(\sqrt{3}\right)^{2}+6^{2}$$

The square of \sqrt{3} is 3.

$$3+6^{2}$$

Calculate 6 to the power of 2 and get 36.

$$3+36$$

Add 3 and 36 to get 39.

$$39$$

Factor

$$3\times 13$$