Question

The volume of a cube is a perfect square. If the side length of the cube is a single digit integer, what will the maximum surface area of the cube (in
m
2
)?​

Answer 1

Answer:

The formula for the volume is given by

(side)3=volume

Since we have the volume, we must take the cube root of the volume to find the length of any one side (since it is a cube, all of the sides are equal).

(side)3−−−−−−√3=side=volume−−−−−−√3

Plugging in 216 for the volume, we end up with

side=6 cm

Answer 2

Answer:

The maximum surface area of the cube is 486 m².

Step-by-step explanation:

• let the side of the cube is a.
• The volume of the cube is $a^{3}$, where a single-digit integer .thus a lies between 1 to 9.
• Given, $a^{3}$ is a perfect square i.e.  $a^{3}=x^2$ where x can be any positive integer.
• Now put the integers from 9 to 1 in the decreasing order to get the value of x. The decreasing order is chosen as the maximum surface area is to be calculated.
• If x=9,

       $9^3=x^2\\x=\sqrt{729}\\ x=27$

• Since 9 satisfies the given condition of the side being the single-digit integer and the  $x^{2}$ being perfectly square.
• The surface area of the square is $6a^2$

         $SA=6*9^2\\SA=486 m^2$