Consider two tasks with different initial data.Task 1.Find the lateral side of the isosceles trapezoidif you know the base BC = b, the base AD = d and the angle at the side of the BAD = alpha.Solution:Drop a perpendicular (the height of the trapezoid) from the vertex B to the intersection with the large basism, will receive a cut BE. Write down the formula for AB using the measure of the angle: AB = AE/cos(BAD) = AE/cos(alpha).
Find the AE. It would be the difference of the lengths of the two bases, divided in half. So: AE = (AD - BC)/2 = (d - b)/2.Now find AB = (d - b)/(2*cos(alpha)).In an isosceles trapezoid the lengths of the sides are equal, hence CD = AB = (d - b)/(2*cos(alpha)).
Task 2.Find the lateral side of the trapezoid AB, if you know the upper base BC = b; lower base AD = d; the height BE = h and the angle opposite the side of the CDA is equal to alpha.Solution:do a second height from vertex C to the intersection with the lower basisof m, will receive a cut of CF. Consider the right triangle CDF, find the direction FD according to the following formula: FD = CD*cos(CDA). The length of the side CD, find another formula: CD = CF/sin(CDA). So: FD = CF*cos(CDA)/sin(CDA). CF = BE = h, therefore, FD = h*cos(alpha)/sin(alpha) = h*ctg(alpha).
Consider the right triangle ABE. Knowing the length of sides AE and be, you can find the third side - the hypotenuse AB. You are aware of the length of the side BE find AE as follows: AE = AD - BC - FD = d - b - h*ctg(alpha).Using the following property of a right triangle - the square of the hypotenuse is equal to the sum of the squares of two - way AB:AB(2) = h(2) + (d - b - h*ctg(alpha))(2).The lateral sides of the trapezoid AB is the square root of the expression on the right side of the equality.
Advice 2: How to find side of a trapezoid
A-line is a regular quadrilateral, having the additional property of parallelism of its two sides, called bases. So the question, first, it should be understood from the point of view of finding the sides. Secondly, for the job of a trapezoid requires at least four parameters.
In this particular case, the General reference (not excessive) should be considered as the condition: the length of the upper and lower bases, and the vector of one of the diagonals. Indices of the coordinates (in order writing formulas was not like multiplication) are shown in italics).For the graphic image of the solution process build figure 1.
Let in the problem is considered a-line AВCD. It contains the lengths of the bases of the armed forces=b and AD=a and the diagonal AC, given as a vector p(px, py). Its length (modulus) |p|=p=sqrt(((px)^2 +(py)^2). Since the vector is set and an angle of inclination to the axis (task - 0X), then label it using f (angle CAD and parallel the angle ACB). Next, you need to apply known to the theorem of cosines. This desired value (the length of a CD or AV when setting up the equation denote by x).
Consider the triangle AСD. Here, the length of side AC is equal to the module of the vector |p|=p. AD=b. By theorem of cosines x^2=p^2+ b^2-2pbcosф. x=CD=sqrt(p^2+ b^2-2pbcosф)=CD.
Now consider the triangle ABC. Length of side AC is equal to the module of the vector |p|=p. BC=a. By theorem of cosines x^2=p^2+ a^2-2расоѕф. x=AB=sqrt(p^2+ a^2-2расоѕф).
Although the quadratic equation has two roots, in this case, you should choose only those where the root of the discriminant of a plus sign, while deliberately excluding negative decisions. This is because the length of the sides of the trapezoid must obviously be positive.
Thus, the required solutions in the form of algorithms solving this problem are obtained. To represent numeric, the decision is to substitute data from the conditions. While Sof is calculated as the direction vector (ORT) vector p=px/sqrt(px^2+py^2).
Of course, other initial data, for example, the specification of the two diagonals and the height of the trapezoid. But in any case you need information about the distance between the bases of the trapezoid.