For the general quadrilateral with sides of lengths a, b, c and d,let s = (a+b+c+d)/2
then area = sqrt[s*(s - a)*(s - b)*(s - c)*(s - d) - a*b*c*d*cos 0.5*(A+C)].
Or if the diagonals are of length p an q and the angle between then is X,
then area = 0.5*p*p*sin(X).
The vertical height can only be used with quadrangles with at least one pair of parallel sides and so is of little use in general.
How do you find the area of an unequal quadrangle?
give me incircle radius of a quadrangle
make it a quadrangle & divide by 2.
There is no opposite for a quadrangle, a typically square area or courtyard between buildings, as at a school.
Multiply half of the sum of the lengths of the top and bottom by half of the sum of the lengths of both sides
You cannot because there is no such thing as a parrallele widths - or even parallel widths.
funge funge
i don't know it this helps, but the sum of the angles in a quadrangle is 360 !
Add the lengths of the four sides
draw a quadrangle below. measure the sides to the nearest 1/2 inch. write the lenght next to each sides. find the perimeter
The sum of the angles in a quadrangle is 360 degrees, whatever the side lengths.
A trapezoid is a quadrangle.