Question

I’m looking at making a trapezium with a box shadow that’s 10px wider at the top than the bottom. In the past I’ve made a trapezium as outlined in the following jsfiddle, but you’ll notice that if I put a box-shadow onto the element it boxes the outerWidth in a rectangle, rather than putting a shadow on the slanted border:

#trapezium {
   margin:20px auto;
   height: 0;
   width: 80px;
   border-bottom: 80px solid blue;
   border-left: 40px solid transparent;
   border-right: 40px solid transparent;
   box-shadow:0 0 10px #333;
}

http://jsfiddle.net/YhePf/8/

My initial thoughts would be to use something along the lines of:

-webkit-transform:perspective(100) rotateX(1deg);

Something like that. While this certainly begins to resolve the issue, I’m not sure what the number 100 refers to in ‘perspective’, and how I could calculate a formula that would make sure the top width was precisely 10px wider than the bottom, regardless of how high or wide this element is.

Any tips? Or a third option to pull this off?

Answer 1

What you’ve built isn’t a trapezoid (aka trapezium) -shaped element; it’s a rectangle-shaped element where the border styling creates the appearance of a trapezoid. This is why the box-shadow is rectangular.

Using the proprietary -webkit-transform property wouldn’t change the shape of the actual element.

To create a truly non-rectangular element, you’ll need to use SVG. See Multi-Shaped CSS Layers \ Non-rectangular CSS Layer or non-rectangular hoverable area.