Solve sqrt{{left(6sqrt{3}right)}^2+{left(4sqrt{3}right)}^2}

$$\sqrt{ { \left(6 \sqrt{ 3 } \right) }^{ 2 } + { \left(4 \sqrt{ 3 } \right) }^{ 2 } }$$

Evaluate

$$2\sqrt{39}\approx 12.489995997$$

Solution Steps

Expand \left(6\sqrt{3}\right)^{2}.

$$\sqrt{6^{2}\left(\sqrt{3}\right)^{2}+\left(4\sqrt{3}\right)^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\sqrt{36\left(\sqrt{3}\right)^{2}+\left(4\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{36\times 3+\left(4\sqrt{3}\right)^{2}}$$

Multiply 36 and 3 to get 108.

$$\sqrt{108+\left(4\sqrt{3}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\sqrt{108+4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\sqrt{108+16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{108+16\times 3}$$

Multiply 16 and 3 to get 48.

$$\sqrt{108+48}$$

Add 108 and 48 to get 156.

$$\sqrt{156}$$

Factor 156=2^{2}\times 39. Rewrite the square root of the product \sqrt{2^{2}\times 39} as the product of square roots \sqrt{2^{2}}\sqrt{39}. Take the square root of 2^{2}.

$$2\sqrt{39}$$