Solve 52^2+sqrt{45}*12^2/27^2

$${ 52 }^{ 2 } + \sqrt{ 45 } \times \frac{ { 12 }^{ 2 } }{ { 27 }^{ 2 } }$$

Evaluate

$$\frac{16\sqrt{5}}{27}+2704\approx 2705.32507732$$

Solution Steps

Calculate 52 to the power of 2 and get 2704.

$$2704+\sqrt{45}\times \frac{12^{2}}{27^{2}}$$

Factor 45=3^{2}\times 5. Rewrite the square root of the product \sqrt{3^{2}\times 5} as the product of square roots \sqrt{3^{2}}\sqrt{5}. Take the square root of 3^{2}.

$$2704+3\sqrt{5}\times \frac{12^{2}}{27^{2}}$$

Calculate 12 to the power of 2 and get 144.

$$2704+3\sqrt{5}\times \frac{144}{27^{2}}$$

Calculate 27 to the power of 2 and get 729.

$$2704+3\sqrt{5}\times \frac{144}{729}$$

Reduce the fraction \frac{144}{729} to lowest terms by extracting and canceling out 9.

$$2704+3\sqrt{5}\times \frac{16}{81}$$

Express 3\times \frac{16}{81} as a single fraction.

$$2704+\frac{3\times 16}{81}\sqrt{5}$$

Multiply 3 and 16 to get 48.

$$2704+\frac{48}{81}\sqrt{5}$$

Reduce the fraction \frac{48}{81} to lowest terms by extracting and canceling out 3.

$$2704+\frac{16}{27}\sqrt{5}$$

Factor

$$\frac{16 {(\sqrt{5} + 4563)}}{27} = 2705.32507732$$