A cylindrical can has an inner radius of 4 cm and an inner height of 2 cm. A flat, rigid square is entirely enclosed inside the can and the square’s four vertices all touch the interior surfaces or edges of the can. What is the number of square centimeters in the area of the largest possible such square?
If you draw a diagram you see that the flat square is 2 dimensional (such as a piece of paper) while the can is 3d. You see that the radius is 4cm. That means if you draw a right angled triangle starting at the center of the circle of the top of the cylinder (lid) it has a leg of 4 cm and the other leg is 4cm then you will have to find the hypotenuse (using the Pythagorean theorem c^2 = leg^2 + leg^2) and solve for c. C is the length of the square : )
Thanks Ms Pi
You are very welcome : )
I get 32 as answer, where as the given answer is 34.
but the cylinder is only 2cm tall, and the square is entirely enclosed.
??
sorry. wrong outburst.
32 is the area of a square lying flat across the top of the cylinder.
But you have a chance to lean it so that the top edge lies in the top of the cylinder, and the bottom edge lies across the bottom of the cylinder.
To find the largest possible square that can fit inside the cylindrical can, we need to consider the dimensions of the can.
The can is cylindrical with an inner radius of 4 cm and an inner height of 2 cm. Since the square's vertices touch the interior surfaces or edges of the can, the side length of the square will be equal to the smaller of the two dimensions of the can.
Therefore, the side length of the square will be 2 cm.
To calculate the area of the square, we square the side length:
Area of the square = (Side length)^2
= 2^2
= 4 square cm
So, the largest possible square that can fit inside the cylindrical can has an area of 4 square centimeters.