To construct a quadrilateral, given the sides AB = a, BC = b, CD = c, AD = d and area S = u², equal to that of a given square.

A posible construction method is as follows :
From the Bretschneider formula S = √(p-a)(p-b)(p-c)(p-d) - abcd cos²( (A+C}/2 )
we can construct angle A+C = φ as :
u cos²(φ/2) = u((p-a)/a)((p-b)/b)((p-c)/c)((p-d)/d) - u(u/a)(u/b)(u/c)(u/d)
in step by step multiplying u by successive ratios (p-a)/a, then (p-b)/b, (p-c)/c and (p-d)/d
Similarily multiplying u by successive ratios u/a, u/b, u/c, u/d
Then construct the difference = u cos²(φ/2).
cos(φ) is then constructed from : cos(φ) = 2cos²(φ/2) - 1

The cosinus law in triangles ABD and BCD gives :
a² + d² - 2ad cos(A) = b² + c² - 2bc cos(φ - A)
That is : (2bc cos(φ) - 2ad)cos(A) + 2bc sin(φ) sin(A) = b² + c² - a² - d²
Construct P = u cos(φ)(b/u)(c/u) - ad/u, Q = bc/u sin(φ) and M = (b²/u + c²/u - a²/u - d²/u)/2
We get P cos(A) + Q sin(A) = M, then construct θ with P = R cos(θ), Q = R sin(θ).
This gives R cos(θ - A) = M, hence the construction of angle A.

The final construction of the quadrilateral from angle A is then obvious.
We have just to carefully choose which intersection point, in case angle C is > 180°.

The following applet performs this construction.
The draggable points a,b,c,d,u define the data
The construction steps are detailed (not too much..) by the "step" buttons
Point A is draggable to move the drawing and see out of range construction points.

Sorry, your Web browser is not Java compliant

The choice of the right intersection point C at the final step (intersection of circles with centers B and D) is done by a "conditionnal construction" : only one of the two intersection points results into the correct area.


Home Arithmetic Geometric Misc Topics Scripts Games Exercices Mail Version Française Previous topic Next topic