Solve sqrt{(-240)^2+(-208+32)^2+(+26-6)^2}

$$\sqrt { ( - 240 ) ^ { 2 } + ( - 208 + 32 ) ^ { 2 } + ( + 26 - 6 ) ^ { 2 } }$$

Evaluate

$$4\sqrt{5561}\approx 298.288451$$

Solution Steps

Calculate -240 to the power of 2 and get 57600.

$$\sqrt{57600+\left(-208+32\right)^{2}+\left(26-6\right)^{2}}$$

Add -208 and 32 to get -176.

$$\sqrt{57600+\left(-176\right)^{2}+\left(26-6\right)^{2}}$$

Calculate -176 to the power of 2 and get 30976.

$$\sqrt{57600+30976+\left(26-6\right)^{2}}$$

Add 57600 and 30976 to get 88576.

$$\sqrt{88576+\left(26-6\right)^{2}}$$

Subtract 6 from 26 to get 20.

$$\sqrt{88576+20^{2}}$$

Calculate 20 to the power of 2 and get 400.

$$\sqrt{88576+400}$$

Add 88576 and 400 to get 88976.

$$\sqrt{88976}$$

Factor 88976=4^{2}\times 5561. Rewrite the square root of the product \sqrt{4^{2}\times 5561} as the product of square roots \sqrt{4^{2}}\sqrt{5561}. Take the square root of 4^{2}.

$$4\sqrt{5561}$$