The trapeze represents the usual quadrangle having additional property of parallelism of two parties which are called the bases. Therefore this question, first, should be understood in terms of search of sides. Secondly, the task of a trapeze requires not less than four parameters.
Instruction
1. In this specific case its most general task (not superfluous) should be considered a condition: lengths of the top and lower bases and also a vector of one of diagonals are given. Indexes of coordinates (that writing formulas was not similar to multiplication) will be italicized). For the graphic representation of process of the decision construct figure 1.
2. Let in the presented task ABCD trapeze be considered. In it lengths of the bases of BC=b and AD=a and also diagonal the EXPERT set by p vector are given (px, py). Its length (module) |p| =p=sqrt (((px) ^2 + (py) ^2). As the vector is set also by a tilt angle to an axis (in a task - 0X), designate it through f (a corner of CAD and ACB corner parallel to it). Further it is necessary to apply the theorem of cosines, known from the school program. At the same time required size (by drawing up the equation designate lengths of CD or AV through x).
3. Consider ACD triangle. Here length of the party the EXPERT is equal to the module of a vector |p| =p. AD=b. According to the theorem of cosines of x^2=p^2 + b^2-2pbcosф. x=CD=sqrt (p^2 + b^2-2pbcosф) =CD.
4. Now consider a triangle of ABC. Length of the party the EXPERT is equal to the module of a vector |p| =p. BC=a. According to the theorem of cosines of x^2=p^2 + a^2-2pacosф. x =AB=sqrt (p^2 + a^2-2pacosф).
5. Though the quadratic equation also has two roots, in this case it is necessary to choose only where the root from a discriminant is faced by a plus, at the same time having obviously excluded negative decisions. It is caused by the fact that length of the party of a trapeze has to be obviously positive.
6. So, required decisions in the form of algorithms of the solution of this task are received. To submit the numerical decision remains given to substitute from a condition. At this cosf it is calculated as the directing vector (ORT) of a vector of p=px/sqrt(px^2+py^2).