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

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

Evaluate

$$3\sqrt{6}\approx 7.348469228$$

Solution Steps

Calculate 6 to the power of 2 and get 36.

$$\sqrt{36+\left(3\sqrt{2}\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\sqrt{36+3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{36+9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\sqrt{36+9\times 2}$$

Multiply 9 and 2 to get 18.

$$\sqrt{36+18}$$

Add 36 and 18 to get 54.

$$\sqrt{54}$$

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$3\sqrt{6}$$