Question

The room has a width of ​8 feet, a length of 12 feet, and a height of 9 feet. ​ ​What is the greatest ​possible distance between two points on the walls in the room?

Answer 1

The greatest possible distance between two points on the walls in the room is 17 feet.

Explanation:

The room can be modelled as a rectangular prism with the following dimensions:

• width = 8 ft
• length = 12 ft
• height = 9 ft

The greatest possible distance between two points on the walls in the room is the line between diagonally opposite corners. This is marked as the blue line (x) on the attached 3D diagram.

This distance is the hypotenuse of a right triangle with a base of the diagonal of the floor and a height of the height of the room.

The diagonal of the floor (marked "y" on the attached diagram) is the hypotenuse of a right triangle with legs of the width and length of the room. To calculate the diagonal of the floor, use Pythagoras Theorem.

`\boxed{\begin{minipage}{9 cm}\underline{Pythagoras Theorem} \\\\$a^2+b^2=c^2$\\\\where:\\ \phantom{ww}$\bullet$ $a$ and $b$ are the legs of the right triangle. \\ \phantom{ww}$\bullet$ $c$ is the hypotenuse (longest side) of the right triangle.\\\end{minipage}}`

Substitute a = 8, b = 12 and c = y into the formula and solve for y:

`\implies 8^2+12^2=y^2`

`\implies 64+144=y^2`

`\implies y^2=208`

`\implies y=√(208)\; \sf ft`

Therefore, the diagonal of the floor (y) is √(208) feet in length.

We can now use Pythagoras Theorem again to find the length x.

Substitute a = y, b = 9 and c = x into the formula:

`\implies y^2+9^2=x^2`

Substitute the found value of y and solve for x:

`\implies (√(208))^2+9^2=x^2`

`\implies 208+81=x^2`

`\implies 289=x^2`

`\implies x=√(289)`

`\implies x=17 \sf \; ft`

Therefore, the greatest possible distance between two points on the walls in the room is 17 feet.