Lesson Trapezoids and their mid-lines

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Trapezoids and their mid-lines


In this lesson you will learn major definitions and facts related to  trapezoids  and their  mid-lines.

Trapezoid  is a quadrilateral which has two opposite sides parallel and the other two sides non-parallel  (see  Figure 1).

The parallel sides of a trapezoid are called its  bases  (sides  AB  and  DC  in  Figure 1).
The non-parallel sides of a trapezoid are called its  lateral sides  or  legs  (sides  AD  and  BC  in  Figure 1).

Mid-line  of a trapezoid is the line segment connecting the midpoints of the lateral sides of a trapezoid.
The mid-line  EF  of the trapezoid  ABCD  is shown in  Figure 2.


    
                    Figure 1.  Trapezoid

            
              Figure 2.  Trapezoid and its mid-line


Theorem 1

The mid-line of a trapezoid is parallel to its bases.
The length of the mid-line of a trapezoid is half of the sum of the lengths of its bases.

Proof

Let  ABCD  be a trapezoid with the bases  AB  and  DC  and the mid-line  EF 
(Figure 2).  Let us draw the straight line  DF  through the points  D  and  F
till the intersection with the extension of the straight line  AB  at the point
G  (Figure 3).  Compare the triangles  DFC  and  FBG.

The segments  FC  and  BF  are congruent since the point  F  is the midpoint
of the side  BC.  The angles  DFC  and  BFG are congruent as the vertical angles.
The angles  DCF  and  FBG are congruent as the alternate exterior angles
at the parallel lines  AB  and  DC  and the transverse  BC  (see the lesson
Parallel lines  under the topic  Angles, complementary, supplementary angles

           
                      Figure 3.  To the proof of the Theorem 1
of the section  Geometry  in this site).

Hence, the triangles  DFC  and  FBG are congruent in accordance with the  ASA-test of congruency of triangles  (see the  Postulate 2 
of the lesson  Congruence tests for triangles  under the topic  Triangles  of the section  Geometry  in this site).

It implies that the segments  DF  and  GF  are congruent as the corresponding sides of the congruent triangles  DFC  and  FBG.

Thus the mid-line  EF  of the trapezoid  ABCD  is the straight line segment connecting the midpoints of the triangle  AGD.

It is well known fact that the the straight line segment connecting the midpoints of the triangle  AGD  is parallel to the triangle base  AG  and its length is half of the length of the triangle base.  See the lesson  The line segment joining the midpoints of two sides of a triangle  under the topic  Triangles  of the section  Geometry  in this site.
In our case,  the length of the segment  EF  is half of the length  AG :   |EF| = 1%2F2*|AG| = 1%2F2*(|AB| + |BG|).
Since  |BG| = |DC|  from the triangles congruency,  we have  |EF| = 1%2F2*(|AB| + |DC|),  or  |EF| = 1%2F2*(a + d),  where  a  and  d  are the lengths of the trapezoid bases.

Thus the proof of the  Theorem 1  is fully completed.


Theorem 2

In a trapezoid,  the line segment drawn from the midpoint of the lateral side parallel to the bases intersects the other lateral side at its midpoint.

Proof

The proof is very close to that of the Theorem 1 above.

Let  ABCD  be a trapezoid with the bases  AB  and  CD,  and let  EF  be the
straight line drawn through the midpoint  E  of the lateral side  AD  parallel
to the bases  (Figure 4).  Let us draw the straight line  DF  through the
points  D  and  F till the intersection with the extension of the straight line
 AB  at the point G  (Figure 4).  Compare the triangles  DFC  and  FBG.

Since the straight line  EF  passes through the midpoint  E  of the triangle
AGD  and is parallel to its base  AB,  it intersects  the other side  GD  of
the triangle at its midpoint  F  too  (see the lesson  The line segment
joining the midpoints of two sides of a triangle  under the topic  Triangles
of the section  Geometry  in this site).  Hence,  the segments  DF  and  FG 

               
                          Figure 4.  To the proof of the Theorem 2
are congruent. Thus the triangles  DFC  and  FBG  have the congruent sides  DF  and  FG.

Further,  the angles  DFC  and  BFG are congruent as the vertical angles.
The angles  FDC  and  FGB are congruent as the alternate interior angles at the parallel lines  AG  and  DC  and the transverse  DG 
(see the lesson  Parallel lines  under the topic  Angles, complementary, supplementary angles  of the section  Geometry  in this site).

Hence, the triangles  DFC  and  FBG are congruent in accordance with the  ASA-test of congruency of triangles  (see the  Postulate 2 
of the lesson  Congruence tests for triangles  under the topic  Triangles  of the section  Geometry  in this site).

It implies that the segments  CF  and  FB  are congruent as the corresponding sides of the congruent triangles  DFC  and  FBG.

Thus the point  F  is the midpoint of the side  BC  of the trapezoid  ABCD.  This is what has to be proved.


Summary

1.  The mid-line of a trapezoid is parallel to its bases.
     The length of the mid-line of a trapezoid is half of the sum of the lengths of its bases.

2.  In a trapezoid,  the line segment drawn from the midpoint of the lateral side parallel to the bases intersects the other lateral side at its midpoint.


Example 1

In a trapezoid the bases are of  17 cm  and  13 cm.  Find the length of the trapezoid's mid-line.

Solution
The mid-line of the trapezoid is half of the sum of the lengths of its bases,  i.e.  %2813+%2B+17%29%2F2 = 30%2F2 = 15 cm.

Answer.  The length of the trapezoid's mid-line is  15 cm.


Example 2

In a trapezoid,  the larger base is of  27 cm long,  and it is in  10 cm  longer than the shorter base.  Find the length of the trapezoid's mid-line.

Solution
The shorter base length is  27 cm - 10 cm = 17 cm.
The mid-line of the trapezoid is half of the sum of the lengths of its bases,  i.e.  %2817+%2B+27%29%2F2 = 44%2F2 = 22 cm.

Answer.  The length of the trapezoid's mid-line is  22 cm.


Example 3

In a trapezoid,  the larger base is in  10 cm longer than the shorter base,  and its mid-line is of  22 cm  long.  Find the lengths of the trapezoid's bases.

Solution
Let x be the length of the larger base of the trapezoid in centimeters.
Then the shorter base length is  (x-10) cm.
Since the mid-line of the trapezoid is half of the sum of the lengths of its bases,  i.e.  %28x+%2B+%28x-10%29%29%2F2 cm,  it gives the equation
%28x+%2B+%28x-10%29%29%2F2 = 22.

Simplify this equation step by step and get the solution:

%282x+-+10%29%2F2 = 22,

2x+-+10 = 44,

2x = 54,

x = 27.

So,  the length of the larger base is  27 cm.  The length of the shorter base is  27 cm - 10 cm = 17 cm.

Answer.  The length of the trapezoid's larger base is  27 cm,  the length of the shorter base is  17 cm.


My other lessons on trapezoids in this site are
    -Trapezoids and their base angles,
    -Diagonals of an isosceles trapezoid are congruent,
    -Mid-line of a trapezoid is the locus of points equidistant from its bases,
    -Solving problems on trapezoids,
    -Solving problems on isosceles trapezoids,
    -Trapezoid is uniquely defined by the lengths of its sides,
    -HOW TO construct a trapezoid given by the lengths of its sides   and
    -PROPERTIES OF TRAPEZOIDS
under the current topic,   and
    -Solved problems on trapezoids   and
    -Solved problems on isosceles trapezoids
under the topic  Geometry  of the section  Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

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