Solve sqrt{left(240+12right)^2+left(-208+40right)^2+left(26-5right)^2}

$$\sqrt{ { \left(240+12 \right) }^{ 2 } + { \left(-208+40 \right) }^{ 2 } + { \left(26-5 \right) }^{ 2 } }$$

Evaluate

$$21\sqrt{209}\approx 303.593478191$$

Solution Steps

Add 240 and 12 to get 252.

$$\sqrt{252^{2}+\left(-208+40\right)^{2}+\left(26-5\right)^{2}}$$

Calculate 252 to the power of 2 and get 63504.

$$\sqrt{63504+\left(-208+40\right)^{2}+\left(26-5\right)^{2}}$$

Add -208 and 40 to get -168.

$$\sqrt{63504+\left(-168\right)^{2}+\left(26-5\right)^{2}}$$

Calculate -168 to the power of 2 and get 28224.

$$\sqrt{63504+28224+\left(26-5\right)^{2}}$$

Add 63504 and 28224 to get 91728.

$$\sqrt{91728+\left(26-5\right)^{2}}$$

Subtract 5 from 26 to get 21.

$$\sqrt{91728+21^{2}}$$

Calculate 21 to the power of 2 and get 441.

$$\sqrt{91728+441}$$

Add 91728 and 441 to get 92169.

$$\sqrt{92169}$$

Factor 92169=21^{2}\times 209. Rewrite the square root of the product \sqrt{21^{2}\times 209} as the product of square roots \sqrt{21^{2}}\sqrt{209}. Take the square root of 21^{2}.

$$21\sqrt{209}$$