hi i have this:

circle (x+4)^2 + (y+4)^2 = 20crosses the x-axis at the points A and B and y-axis at points C and D. Find the area of the quadrilateral ABCD.

I spend so much time trying to figure height etc. but all i could get was sides lenghts, so the only way i could get to the solution was by using this quadrilateral formula
x1 * y2 - y1 * x2 .... but as am doing this from textbook that way is never mentioned anywhere , so can someone please tell what would be the other way of doing it ?

$(x+4)^2 + (0+4)^2 = 20 \implies x=-6 \text{ and } x=-2$

$x=-6 \implies 4 + (y+4)^2 = 20 \implies y=0 \text{ and } y=-8$

rectangle base = $|-2 - (-6)| = 4$

rectangle height = $|0 - (-8)| = 8$

Sometimes quadrilateral means that the polygon is convex. I prefer to say a quadrilateral is any polygon with 4 vertices, so edges can intersect at points other than the vertices. So there can be two quadrilaterals ABCD described by your problem, depending on how the vertices are labeled. Idea has given you one of these. The other is drawn below. Sometimes, it is helpful to know the answers. The convex quadrilateral has area 16 and the other quadrilateral has area 6. For either of these, it is helpful to have the formula for the distance from a point $(x_0,y_0)$ to a line with equation $Ax+By+C=0$, namely
$${Ax_0+By_0+C\over \sqrt{A^2+B^2}}$$
This helps to find heights.

Edit: By the way, the formula for area that you mentioned is valid only for convex quadrilaterals.

well ... looks like I had rectangle on the brain. nevermind.

Originally Posted by johng
Sometimes quadrilateral means that the polygon is convex...
Basically im not on this level yet.. its the trapezoid in my case. But trying everything i got a lot similiar values to 16 , they just differ by decimal places. only the formula i mentioned before gave me exact 16 , not sure why.
On the other note, could i just calculate bases and then calculate the 2 triangles on sides and by doing that i autamatically get trapezs height right ?

A convex quadrilateral is just a quadrilateral Q such that for any 2 points "inside" Q, the line segment joining the 2 points lies entirely inside Q. The first quadrilateral is convex.

Now this quad is not convex; choose 2 points inside of the 2 triangles. The line connecting the 2 points does not stay inside the quad. You do the computations as indicated. If you have trouble, post again.

Originally Posted by johng
Now this quad is not convex; choose 2 points inside of the 2 triangles. The line connecting the 2 points does not stay inside the quad. You do the computations as indicated. If you have trouble, post again.
@johng, I don't know anyone actively involved in axiomatic geometry who thinks of your second figure as a quadrilateral. In fact going back to Hilbert, E H Moore, R L Moore, the four segments were required to be non-intersecting. It is true that some modern authors have defined a figure called variously a cross-quadrilateral, crossed quadrilateral, butterfly quadrilateral or bow-tie quadrilateral.

Moreover, a careful reading of the original question gives $A,~B,~C,~\&~D$ in that particular order. Those in the editing community would expect the test taker to use $A: (-6,0),~B: (-2,0),~C: (0,-2),~\&~D: (0,-6)$ to the quadrilateral $ABCD$.