Re: 3D figure cut by a plane

Have you tried sketching it? Also, the solid has more than one surface so we should be a bit more specific.

Re: 3D figure cut by a plane

It is a hexagon. How do I prove this mathematically...?????

Re: 3D figure cut by a plane

Re: 3D figure cut by a plane

Not sure what tools you have available to you, but the simplest way is to imagine the side of the cube is 2, then you can take two vectors (1,1,0) and (0,1,-1) and find the angle between them using the dot product. It will come out to 120 degrees and the symmetry should not require you to work out other angles. If you do not have vectors in your locker then you will have to play around with Pythagoras I think.

Re: 3D figure cut by a plane

Hello, Aditya3003!

If the answer is *hexagon*, there is a typo.

. . Please correct it.

Quote:

There is a cube with base $\displaystyle ABCD$ and top $\displaystyle PQRS.$

A plane cuts this cube and passes through the mid-points of $\displaystyle DC, CB$ and $\displaystyle {\color{blue}RB}.\;\;{\color{red}??}$

What shape is formed on the plane ?

Assuming $\displaystyle P$ is above $\displaystyle A$, $\displaystyle Q$ is above $\displaystyle B$, etc.

. . the diagram looks like this:

Code:

` P*---------*S`

/: /|

/ : / |

/ : / |

Q*---+-----*R |

| : | |

| A* - - + - *D

| / | /

| / | o M

|/ |/

B*----o----*C

N

$\displaystyle M$ is the midpoint of $\displaystyle DC$; $\displaystyle N$ is the midpoint of $\displaystyle CB.$

But $\displaystyle RB$ is a *diagonal* of the front face.

The third midpoint is $\displaystyle O$, the center of square $\displaystyle BCRQ$.

The shape is right triangle $\displaystyle MNO.$

Re: 3D figure cut by a plane

I also made the assumption initially that P is above A. That would seem logical right? Well looking at the image he attached, it seems that the assumption is wrong. So given the way he has drawn it, we do get a hexagon.

Re: 3D figure cut by a plane

According to your image midpoint of DS. Sorry for the carelessness...