Question: What would cyclopropane look like if it weren't planar?

How and why does non-planar cyclopropane have higher torsional strain than planar cyclopropane? can you please show me visually via image that non-planar cyclopropane has higher torisional strain and maybe more angle strain than planar cyclopropane? 

On the otherhand, why and how is cyclobutane bulged or puckering, a.k.a non-planar?

Answer: 
> 
> Why is cyclopropane planar?
> 
> 
> 

This is like asking why $\ce{HCl}$ is linear–just as two points define a line, three points define a plane. There isn't any way to orient the three methylene groups in space that they aren't coplanar.

$\hspace{8cm}$[![plane](https://i.stack.imgur.com/QGEHjs.png)](https://i.stack.imgur.com/QGEHjs.png)

Torsional strain in cyclopropane is also significant for this reason (the molecule has a $\mathrm{C\_3}$ axis through the center of the ring, meaning every hydrogen is overlapping). Any reduction in torsional strain, however, decreases the already weak carbon-carbon $\ce{sp^3}$ orbital overlap. Though the $\ce{C-C-C}$ bond angles are $60º$, the orbitals are still oriented roughly $109.5º$ from one another, resulting in very weak and highly strained [banana bonds](https://en.wikipedia.org/wiki/Bent_bond).

$\hspace{8cm}$[![strain](https://i.stack.imgur.com/MiWWNs.png)](https://i.stack.imgur.com/MiWWNs.png)

> 
> Why is cyclobutane nonplanar?
> 
> 
> 

Cyclobutane has one additional methylene, and can reduce its torsional strain by orienting one methylene out of the plane of the other three.

$\hspace{7.3cm}$[![cyclobutane](https://i.stack.imgur.com/Z7oIG.png)](https://i.stack.imgur.com/Z7oIG.png)

This is in part possible due to the much greater carbon-carbon $\ce{sp^3}$ orbital overlap.
If cyclopropane did pucker then a new plane would exist. This is because any set of three points (a.k.a three carbons) is coplanar. Cyclobutane puckering can occur and make a difference as a set of four points/carbons is not necessarily coplanar.