Solve {left(frac{40sqrt{3}}{3}right)}^2+40^2

$${ \left( \frac{ 40 \sqrt{ 3 } }{ 3 } \right) }^{ 2 } + { 40 }^{ 2 }$$

Evaluate

$$\frac{6400}{3}\approx 2133.333333333$$

Solution Steps

To raise \frac{40\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(40\sqrt{3}\right)^{2}}{3^{2}}+40^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$\frac{\left(40\sqrt{3}\right)^{2}}{3^{2}}+1600$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1600 times \frac{3^{2}}{3^{2}}.

$$\frac{\left(40\sqrt{3}\right)^{2}}{3^{2}}+\frac{1600\times 3^{2}}{3^{2}}$$

Since \frac{\left(40\sqrt{3}\right)^{2}}{3^{2}} and \frac{1600\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(40\sqrt{3}\right)^{2}+1600\times 3^{2}}{3^{2}}$$

Expand \left(40\sqrt{3}\right)^{2}.

$$\frac{40^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}+1600$$

Calculate 40 to the power of 2 and get 1600.

$$\frac{1600\left(\sqrt{3}\right)^{2}}{3^{2}}+1600$$

The square of \sqrt{3} is 3.

$$\frac{1600\times 3}{3^{2}}+1600$$

Multiply 1600 and 3 to get 4800.

$$\frac{4800}{3^{2}}+1600$$

Calculate 3 to the power of 2 and get 9.

$$\frac{4800}{9}+1600$$

Reduce the fraction \frac{4800}{9} to lowest terms by extracting and canceling out 3.

$$\frac{1600}{3}+1600$$

Add \frac{1600}{3} and 1600 to get \frac{6400}{3}.

$$\frac{6400}{3}$$

Factor

$$\frac{2 ^ {8} \cdot 5 ^ {2}}{3} = 2133\frac{1}{3} = 2133.3333333333335$$