Day 10: Diagonals of a Cuboid

Geometry Level 2

The three face diagonals (shown in the diagram) of a cuboid are of length $\sqrt{55}, 8$ and $9$ .

Find the length of the longest diagonal in the cuboid.

This problem is part of the Advent Calendar 2015 .

The answer is 10.

1 solution

Michael Ng
Dec 9, 2015

Letting the sides of the cuboid be $a,b$ and $c$ , by Pythagoras' Theorem we have $a^2+b^2 = 55 \\ b^2+c^2 = 64 \\ c^2 + a^2 = 81$ Adding these together gives $2(a^2+b^2+c^2) = 200$ and so $a^2+b^2+c^2 = 100$ .

You can show that the length of the longest diagonal is equal to $\sqrt{a^2+b^2+c^2}$ , so the answer is $\sqrt{100} = \boxed{10}$ , as required.

To be complete we must show that such a cuboid is possible; solving the equations shows that a cuboid with side lengths $\sqrt{45}, 6, \sqrt{19}$ works.

Same way.. And nice problem

Dev Sharma - 5 years, 6 months ago

Thanks, I really appreciate it!

Michael Ng - 5 years, 6 months ago

Nice problem and nice solution.

Anupam Nayak - 5 years, 6 months ago

Thank you!

Michael Ng - 5 years, 6 months ago

As long as triangle { $\sqrt{55}$ , 8 and 9} is possible, is there any impossible cuboid?

Lu Chee Ket - 5 years, 6 months ago

Nice question! I think the answer is no; I should have phrased my solution as the only cuboid that works is the one given above.

Because we have the value of $a^2+b ^2+c^2$ , then we can take away each of equations from this value giving the values of $a^2, b^2$ and $c^2$ which will give us our unique cuboid.

Thanks for pointing this out!

Michael Ng - 5 years, 6 months ago

Actually because you found that $a^2$ , $b^2$ and $c^2$ are easy to derive from equations we have, you solved and certain about a valid unique cuboid, all right!

Lu Chee Ket - 5 years, 6 months ago

Day 10: Diagonals of a Cuboid

This problem is part of the Advent Calendar 2015 .

The answer is 10.

1 solution

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