Instruction

1

Draw a rectangular trapezoid ABCD. The sides of this figure indicate, respectively, as AB and DC. The first side DC coincides with the height of the trapezoid. It is perpendicular to two rectangular bases of the trapezoid.

There are several ways of finding the sides. So for example, if the task is given to the second lateral side BA and the angle ABH=60, then the height of the first find the most simple of ways, having a height of BH:

BH=AB*sinα

Since BH=CD, then CD=AB*sinα=√3AB/2

There are several ways of finding the sides. So for example, if the task is given to the second lateral side BA and the angle ABH=60, then the height of the first find the most simple of ways, having a height of BH:

BH=AB*sinα

Since BH=CD, then CD=AB*sinα=√3AB/2

2

If, on the contrary, given of the trapezoid is denoted as CD, and you want to find her the same side of AB, this problem is solved in some other way. Since BH=CD and at the same time, BH is a leg of the triangle ABH, we can conclude that the side AB is equal to:

AB=BH/sinα=2BH/√3

AB=BH/sinα=2BH/√3

3

The problem can be solved and in that case, if the angle is unknown, provided that the two bases and the side AB. However, in this case, you can only find the side CD, which is the height of the trapezoid. Initially, knowing the value of the grounds, find the length of line AH. It is equal to the difference between the larger and the smaller bases, because we know that BH=CD:

AH=AD-BC

Then, using the Pythagorean theorem, find the height BH is equal to the side CD:

BH=√AB^2-AH^2

AH=AD-BC

Then, using the Pythagorean theorem, find the height BH is equal to the side CD:

BH=√AB^2-AH^2

4

If a rectangular trapezoid has the diagonal BD and angle 2α, as shown in figure 2, the side AB can also be found by the Pythagorean theorem. To do this, first calculate the length of base AD:

AD=BD*cos2α

Then find the side AB in the following way:

AB=√BD^2-AD^2

Then prove the similarity of triangles ABD and BCD. Since these triangles share one common side - diagonal, and thus, two angles are equal, as can be seen from the figure, these shapes are similar. Based on this evidence, find the second lateral side. If you know the upper base and the diagonal, you find the normal way using a standard spherical law of cosines:

c^2=a^2+b^2-2ab cos α, where a, b, C be the sidelengths of a triangle, α is the angle between sides a and b.

AD=BD*cos2α

Then find the side AB in the following way:

AB=√BD^2-AD^2

Then prove the similarity of triangles ABD and BCD. Since these triangles share one common side - diagonal, and thus, two angles are equal, as can be seen from the figure, these shapes are similar. Based on this evidence, find the second lateral side. If you know the upper base and the diagonal, you find the normal way using a standard spherical law of cosines:

c^2=a^2+b^2-2ab cos α, where a, b, C be the sidelengths of a triangle, α is the angle between sides a and b.

# 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.

Instruction

1

*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.*

2

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).

3

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.4

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).5

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.6

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).

Note

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.