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 Cuboid Concern (Posted on 2015-05-21)
A rectangular cuboid has side lengths 5, 7 and 10. Its center is O. The plane P passes through O and is perpendicular to one of the space diagonals. Find the area of its intersection with the cuboid.

 No Solution Yet Submitted by K Sengupta No Rating

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 If I didn't make any mistakes along the way | Comment 1 of 2
The space diagonal has length sqrt(174) and a 5 x 10 base has face diagonal length of sqrt(125). The space diagonal is therefore tilted arccos(sqrt(125/174)) to the plane of either 5 x 10 side. So the cutting plane makes the complementary angle arcsin(sqrt(125/174)) with the 5 x 10 side.

Similarly, the cutting plane makes an angle of arcsin(sqrt(149/174)) with the 7 x 10 wall of the box.  And of course the 5x10 top of the box is perpendicular (90°) to the 7x10 wall.

Construct (mentally of course) a sphere about the point where the cutting plane intersects both these walls.

The planes involved cut a spherical triangle on the surface of this sphere, and the dihedral angles make the angles on this sphere, while the sides of the sphere equal the planar angles.  The angle of intersection on the top 5x10 face corresponds to the side of the spherical triangle between the 90° angle and the arcsin(sqrt(125/174)) angle.  The angle of intersection on the top 7x10 face corresponds to the side of the spherical triangle between the 90° angle and the arcsin(sqrt(149/174)) angle. The side of the spherical triangle opposite the 90° angle represents the acute angle of the parallelogram whose area is sought.

To find the sides of the spherical triangle we can use one of the spherical laws of cosines: cos(A) = -cos(B)cos(C) + sin(B)sin(C)cos(a):

cos(a)= (cos(A)+cos(B)cos(C)) / (sin(B)sin(C))

cos(90) = 0
cos(arcsin(sqrt(125/174))) = sqrt(49/174)
cos(arcsin(sqrt(149/174))) = sqrt(25/174)

The face angle between the cutting plane and the 5x10 top is then:

arccos((sqrt(25/174)+0)/(sqrt(125/174)*1) = arccos(sqrt(1/5))
= arcsin(sqrt(4/5))

The top side of the parallelogram (along the 5x10 face) is therefore 5/sqrt(4/5).

The face angle between the cutting plane and the 7x10 side is then:

arccos((sqrt(49/174)+0)/(sqrt(149/174)*1) = arccos(sqrt(49/149))
= arcsin(sqrt(100/149))

The other side of the parallelogram (along the 7x10 face) is therefore 7/sqrt(100/149).

The planar angle between these two sides of the parallelogram is given by:

arccos(0 + sqrt(49*25)/174) / (sqrt(125*149)/174))

= arccos(35/sqrt(125*149))

To get the area of the parallelogram we need to multiply the two side lengths and then multiply by sin(arccos(35/sqrt(125*149))):

(5/sqrt(4/5))*(7/sqrt(100/149))*sin(arccos(35/sqrt(125*149)))

Plugging this into a calculator gives 46.1681708539551 as that level of approximation.

There was a lot of room for error above, but the answer is reasonably larger than the 35 from cutting straight across.

I'm making the assumption here, that I didn't check, that the cutting plane does not intersect the 5x7 end walls.

 Posted by Charlie on 2015-05-21 16:16:16

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