HOW DO YOU SEE IT? The edge length $s$ of a cube is an irrational number, the surface area is an irrational number, and the volume is a rational number. Which could be $s$ ? $\newline$ $\frac{2}{3}$ $\newline$ $\pi$ $\newline$ $\sqrt{2}$ $\newline$ $\sqrt[3]{2}$

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Q. HOW DO YOU SEE IT? The edge length

s

of a cube is an irrational number, the surface area is an irrational number, and the volume is a rational number. Which could be

s

\newline

\frac{2}{3}

\newline

\pi

\newline

\sqrt{2}

\newline

\sqrt[3]{2}

Analyze Cube Properties: Let's analyze the properties of a cube. The surface area $A$ of a cube is given by the formula $A = 6s^2$ , where $s$ is the edge length. The volume $V$ of a cube is given by the formula $V = s^3$ . We are looking for an edge length $s$ such that the surface area is irrational and the volume is rational.
Option $1$ : $s = \frac{2}{3}$ : Let's consider the first option, $s = \frac{2}{3}$ . If $s$ is $\frac{2}{3}$ , then the surface area $A = 6 \times \left(\frac{2}{3}\right)^2 = 6 \times \frac{4}{9} = \frac{24}{9} = \frac{8}{3}$ , which is a rational number. The volume $V = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$ , which is also a rational number. This does not meet the criteria since the surface area is not irrational.
Option $2$ : $s = \pi$ : Now let's consider the second option, $s = \pi$ . If $s$ is $\pi$ , then the surface area $A = 6 \times \pi^2$ , which is an irrational number because $\pi$ is irrational and the square of an irrational number is also irrational. The volume $V = \pi^3$ , which is also an irrational number. This does not meet the criteria since the volume is not rational.
Option $3$ : $s = \sqrt{2}$ : Next, let's consider the third option, $s = \sqrt{2}$ . If $s$ is $\sqrt{2}$ , then the surface area $A = 6 \times (\sqrt{2})^2 = 6 \times 2 = 12$ , which is a rational number. The volume $V = (\sqrt{2})^3 = 2\sqrt{2}$ , which is an irrational number. This does not meet the criteria since the surface area is not irrational.
Option $4$ : $s = \sqrt[3]{2}$ : Finally, let's consider the fourth option, $s = \sqrt[3]{2}$ . If $s$ is the cube root of $2$ , then the surface area $A = 6 \times (\sqrt[3]{2})^2$ . Since the square of the cube root of $2$ is not a perfect square, it is an irrational number. The volume $V = (\sqrt[3]{2})^3 = 2$ , which is a rational number. This meets the criteria since the surface area is irrational and the volume is rational.