Solve sqrt{20^2+(16sqrt{71})^2}

$$\sqrt { 20 ^ { 2 } + ( 16 \sqrt { 71 } ) ^ { 2 } }$$

Evaluate

$$12\sqrt{129}\approx 136.293800299$$

Solution Steps

Calculate 20 to the power of 2 and get 400.

$$\sqrt{400+\left(16\sqrt{71}\right)^{2}}$$

Expand \left(16\sqrt{71}\right)^{2}.

$$\sqrt{400+16^{2}\left(\sqrt{71}\right)^{2}}$$

Calculate 16 to the power of 2 and get 256.

$$\sqrt{400+256\left(\sqrt{71}\right)^{2}}$$

The square of \sqrt{71} is 71.

$$\sqrt{400+256\times 71}$$

Multiply 256 and 71 to get 18176.

$$\sqrt{400+18176}$$

Add 400 and 18176 to get 18576.

$$\sqrt{18576}$$

Factor 18576=12^{2}\times 129. Rewrite the square root of the product \sqrt{12^{2}\times 129} as the product of square roots \sqrt{12^{2}}\sqrt{129}. Take the square root of 12^{2}.

$$12\sqrt{129}$$