Right trapezoid properties, relationships and formulas, examples

A rectangle trapezoid is a flat figure with four sides, such that two of them are parallel to each other, called bases and also one of the other sides is perpendicular to the bases.

For this reason, two of the internal angles are right, that is, they measure 90º. Hence the name "rectangle" given to the figure. The following image of a right trapezoid clarifies these characteristics:

Article index

• 1 Elements of the trapezoid
• 2 Relations and formulas
  • 2.1 Height h of the trapezoid
  • 2.2 Perimeter P
  • 2.3 Average base
  • 2.4 Area
  • 2.5 Diagonals, sides and angles
• 3 Examples of right trapezoids
  • 3.1 The trapezoid as a design element
  • 3.2 Trapezoidal wave generator
  • 3.3 In numerical calculation
  • 3.4 Beam with trapezoidal load
  • 3.5 As an educational and learning tool
• 4 Solved exercises
  • 4.1 - Exercise 1
  • 4.2 - Exercise 2
• 5 References

Elements of the trapezoid

The elements of the trapezoid are:

-Bases

-Vertices

-Height

-Internal angles

-Middle base

-Diagonals

We are going to detail these elements with the help of figures 1 and 2:

The sides of the right trapezoid are denoted by lowercase letters a, b, c, and d. The corners of the figure o vertices They are indicated in capital letters. Finally the internal angles They are expressed in Greek letters.

By definition, bases of this trapezoid are sides a and b, which, as can be seen, are parallel and also have different lengths.

The side perpendicular to both bases is the side c to the left, which is the height h of the trapeze. And finally there is the side d, which forms the acute angle α with the side a.

The sum of the internal angles of a quadrilateral is 360º. It is easily appreciated that the missing angle C in the figure is 180 - α.

The middle base is the segment joining the midpoints of the non-parallel sides (segment EF in figure 2).

And finally there are the diagonals d₁ and d_two, the segments that join the opposite vertices and that intersect at point O (see figure 2).

Relationships and formulas

Trapezoid height h

h = c

Perimeter P

It is the measure of the contour and is calculated by adding the sides:

Perimeter = a + b + c + d

The side d is expressed in terms of the height or the side c using the Pythagorean theorem:

d = √ (a-b)^two + c^two

Substituting in the perimeter:

P = a + b + c + √ (a-b)^two + c^two

Middle base

It is the semi-sum of the bases:

Mean base = (a + b) / 2

Sometimes the mean base is found expressed in this way:

Average base = (Major base + minor base) / 2

Area

The area A of the trapezoid is the product of the mean base times the height:

A = (Major base + minor base) x height / 2

A = (a + b) c / 2

Diagonals, sides and angles

Several triangles appear in Figure 2, both right and non-right. The Pythagorean theorem can be applied to those that are right triangles and to those that are not, the cosine and sine theorems.

In this way relationships are found between the sides and between the sides and internal angles of the trapezoid..

CPA triangle

It is a rectangle, its legs are equal and are worth b, while the hypotenuse is the diagonal d₁, Thus:

d₁^two = b^two + b^two = 2b^two

DAB triangle

It is also a rectangle, the legs are to Y c (or also to Y h) and the hypotenuse is d_two, so that:

d_two^two = a^two + c^two = a^two + h^two

CDA triangle

Since this triangle is not a right triangle, the cosine theorem is applied to it, or also the sine theorem.

According to the cosine theorem:

d₁^two = a^two + d^two - 2ad cos α

CDP triangle

This triangle is a right triangle and with its sides the trigonometric ratios of the angle α are constructed:

sin α = h / d

cos α = PD / d

But the side PD = a - b, therefore:

cos α = (a-b) / d → a - b = d cos α

a = b + d cos α

You also have:

tg α = sin α / cos α = h / (a-b) → h = tg α (a-b)

CBD triangle

In this triangle we have the angle whose vertex is at C. It is not marked in the figure, but at the beginning it was highlighted that it is 180 - α. This triangle is not a right triangle, so the cosine theorem or sine theorem can be applied..

Now, it can easily be shown that:

sin (180 - α) = sin α

cos (180 - α) = - cos α

Applying the cosine theorem:

d_two^two = d^two + b^two - 2db cos (180 - α) = d^two + b^two + 2db cos α

Examples of right trapezoids

Trapezoids and in particular right trapezoids are found on many sides, and sometimes not always in tangible form. Here we have several examples:

The trapezoid as a design element

Geometric figures abound in the architecture of many buildings, such as this church in New York, which shows a structure in the shape of a rectangular trapezoid.

Likewise, the trapezoidal shape is frequent in the design of containers, containers, blades (cutter or exact), badges and in graphic design.

Trapezoidal wave generator

Electrical signals can not only be square, sinusoidal or triangular. There are also trapezoidal signals that are useful in many circuits. In figure 4 there is a trapezoidal signal composed of two right trapezoids. Between them they form a single isosceles trapezoid.

In numerical calculation

To calculate in numerical form the definite integral of the function f (x) between a and b, the trapezoid rule is used to approximate the area under the graph of f (x). In the following figure, on the left the integral is approximated with a single right trapezoid.

A better approximation is the one in the right figure, with multiple right trapezoids.

Trapezoidal loaded beam

Forces are not always concentrated on a single point, since the bodies on which they act have appreciable dimensions. Such is the case of a bridge over which vehicles circulate continuously, the water of a swimming pool on the vertical walls of the same or a roof on which water or snow accumulates..

For this reason, forces are distributed per unit length, surface area or volume, depending on the body on which they act..

In the case of a beam, a force distributed per unit length can have various distributions, for example the right trapezoid shown below:

In reality, distributions do not always correspond to regular geometric shapes like this one, but they can be a good approximation in many cases..

As an educational and learning tool

Blocks and pictures with geometric shapes, including trapezoids, are very useful for children to become familiar with the fascinating world of geometry from an early age.

Solved exercises

- Exercise 1

In the right trapezoid in figure 1, the larger base is 50 cm and the smaller base is equal to 30 cm, it is also known that the oblique side is 35 cm. Find:

a) Angle α

b) Height

c) Perimeter

d) Average base

e) Area

f) Diagonals

Solution to

The statement data is summarized as follows:

a = major base = 50 cm

b = smaller base = 30 cm

d = slant side = 35 cm

To find the angle α we visit the formulas and equations section to see which one best suits the data provided. The sought angle is found in several of the analyzed triangles, for example the CDP.

There we have this formula, which contains the unknown and also the data that we know:

cos α = (a-b) / d

Therefore:

α = arcs [(a-b) / d] = arches [(50-30) / 35] = arches 20/35 = 55.15 º

Solution b

From the equation:

sin α = h / d

It clears h:

h = d. sin α = 35 sin 55.15 º cm = 28.72 cm

Solution c

The perimeter is the sum of the sides, and since the height is equal to side c, we have:

c = h = 28.72 cm

Therefore:

P = (50 + 30 + 35 + 28.72) cm = 143.72 cm

Solution d

The mean base is the semi-sum of the bases:

Middle base = (50 + 30 cm) / 2 = 40 cm

Solution e

The area of ​​the trapezoid is:

A = average base x height = 40 cm x 28.72 = 1148.8 cm^two.

Solution f

For the diagonal d₁ you can use this formula:

d₁^two = b^two + b^two = 2b^two

d₁^two= 2 x (30 cm)^two = 1800 cm^two

d₁ = √1800 cm^two = 42.42 cm

And for the diagonal d_two:

d_two^two = d^two + b^two + 2db cos α = (35 cm)^two + (30 cm)^two + 2 x 35 x 30 cm^two cos 55.15 º = 3325 cm^two

d_two = √ 3325 cm^two = 57.66 cm

This is not the only way to find d_two, since there is also the DAB triangle.

- Exercise 2

The following graph of velocity as a function of time belongs to a mobile that has uniformly accelerated rectilinear motion. Calculate the distance traveled by the mobile during the time interval between 0.5 and 1.2 seconds.

Solution

The distance traveled by the mobile is numerically equivalent to the area under the graph, delimited by the indicated time interval.

The shaded area is the area of ​​a right trapezoid, given by:

A = (Major base + minor base) x height / 2

A = (1.2 + 0.7) m / s x (1.2 - 0.5) s / 2 = 0.665 m

References

1. Baldor, A. 2004. Plane and space geometry with trigonometry. Cultural Publications.
2. Bedford, A. 1996. Statics. Addison Wesley Interamericana.
3. Jr. geometry. 2014. Polygons. Lulu Press, Inc.
4. OnlineMSchool. Rectangular trapezoid. Recovered from: es.onlinemschool.com.
5. Automatic geometry problem solver. The trapeze. Recovered from: scuolaelettrica.it
6. Wikipedia. Trapezoid (geometry). Recovered from: es.wikipedia.org.