At a restaurant, a trash can’s opening is rectangular and measures 7 inches by 9 inches. The restaurant serves food on trays that measure 12 inches by 16 inches. Jada says it is impossible for the tray to accidentally fall through the trash can opening because the shortest side of the tray is longer than either edge of the opening. Do you agree or disagree with Jada’s explanation? Explain your reasoning.

The trash can's opening measures 7 inches by 9 inches and the trays measure 12 inches by 16 inches. It's obvious that the trays won't fit through the opening if they are tried by the opening's width or height, but what if we try to make the tray go through the diagonal of the opening?

We must then compute the length of the opening's diagonal, if it's longer than any of the tray's dimensions, then it can fall through it.

[tex]d=\sqrt{7^2+9^2}=11.4\ in[/tex]

The diagonal is shorter than the shortest side of the tray, so it's impossible to fall through the opening, but not for the reason explained by Jada. Some calculation should be done before stating it.

codiepienagoya codiepienagoya · Answer 2 · 2021-11-22T17:42:42+01:00

Following are the solution to the given question:

The garbage can entrance is 7 inches by 9 inches, and the trays are 12 inches by 16 inches.
The trays will not fit through the aperture if we try to force them through the width or height of the opening, but what if we try to force the tray through the diagonal of the opening.
You then must calculate the length of an opening's diagonal; whether it is longer than any of the tray's dimensions, this can drop through all of it.

[tex]d=\sqrt{7^2+9^2}= \sqrt{49+81} = \sqrt{130} =11.4\ in[/tex]

Since the diagonal is shorter than that of the selecting appropriate tray, it's also impossible to fall through the hole, though not for the reason Jada explains. After expressing it, little math must be performed.

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