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What are the types of quadrilaterals?

Quadrilaterals are part of a plane enclosed by four sides (quad means four and lateral means side). All quadrilaterals have exactly four sides and four angles, and they can be sorted into specific groups based on lengths of their sides or measures of their angles. What they have in common is that in every quadrilateral the sum of the measures of all interior angles is equal to 360^{\circ}.

Their vertices are marked with capital letters and sides with small letters.

Angles in vertices A, B, C and D are usually marked in order with: \alpha, \beta, \gamma, \delta (alpha, beta, gamma, delta).

Remember that in triangles, the sum of  the measures of all exterior angles is equal to 360^{\circ} (remember, exterior angle is a supplementary angle to a certain interior angle).

This will also be true for quadrilaterals. The sum of all measures of exterior angles in quadrilaterals is always equal to 360^{\circ}.

Diagonals are lines that connect opposite angles.

The division of quadrilaterals according to perpendicularity diagonals and parallel sides:

 

First group of quadrilaterals is a scalene quadrilateral. Scalene quadrilateral is a quadrilateral that doesn’t have any special properties; the sides  and angles have different lengths and measures.

Quadrilaterals which have one pair of parallel sides are called trapezoids. Sides that are parallel are called bases of a trapezoid, and ones that are not parallel are called legs.

Trapezoids whose legs are of equal length are called isosceles trapezoids.

Diagonals of a isosceles trapezoids are congruent.

Height or altitude of a trapezoid is the length of a line that is perpendicular to a base and runs through opposite vertex. Altitude of a trapezoid will be equal no matter from which vertex we draw it. If we are drawing an altitude from larger base, we simply extend the shorter base.

Theorem.

If \alpha is an angle in vertex A, \beta in vertex B, \gamma in vertex C and \delta in vertex D in a trapeziod ABCD, then is valid:

\alpha + \delta = 180^{\circ}
\beta + \gamma = 180^{\circ}.

In other words, the angles on the same side of a leg of a trapezoid are supplementary.

Proof.

 

Expand the segment \overline{AD} over the vertices A and D. On the line AD denote point E.  Since the line AD is a transverse of the parallel lines AB and CD,  then \angle{CDE} =\alpha is valid . The angles \angle{CDE} and \delta are supplementary angles, which means that  \angle{CDE} +\delta = 180^{\circ} \Rightarrow \alpha+ \delta =  180^{\circ}.

Analogously, we obtained  \beta+ \gamma =  180^{\circ}.

 

A parallelogram is a quadrilateral whose opposite sides are congruent and parallel.

Altitude or a height of a parallelogram, in the label h, is the line segment that connects a vertex with opposite side, and is perpendicular to that side.

Theorem.

Let ABCD be a parallelogram. The opposite angles in a quadrilateral ABCD are congruent, and the adjacent angles  are supplementary.

Proof.

By definition,  if  ABCD  is a trapezoid with legs \overline{AD} and \overline{BC}, then:

    \[\beta + \gamma = 180^{\circ}  \quad and  \quad \alpha + \delta = 180^{\circ}.\]

If  ABCD  is a trapezoid with legs \overline{AB} and \overline{CD}, then:

    \[\alpha + \beta = 180^{\circ}  \quad and  \quad \beta + \delta = 180^{\circ}.\]

It follows \beta = \delta and \alpha = \gamma.

 

Theorem.

The following statements are equivalent to each other:

1) A quadrilateral ABCD is a parallelogram

2) There exists two opposite sides of a quadrilateral ABCD which are congruent and parallel

3) Each two opposites sides of a quadrilateral ABCD are congruent

4) Diagonals of a quadrilateral ABCD bisect each other

5) Both pairs of opposite angles of a quadrilateral ABCD are congruent

Each of the above statements can be an alternative definition of a parallelogram. The remaining statements we need to prove.

Proof.

1) \Rightarrow 2)

Let ABCD be a parallelogram. Then \overline{AB} \| \overline{CD} and \overline{AD} \| \overline{BC}.

 

 

Since line AC is a traverse of parallel lines AB and CD ,then \angle{ACD} =\angle{CAB}. A line AC is also a traverse of parallel lines BC and AD that’s \angle{ACB}=\angle{DAC}.

\overline{AC} is also the common side of triangles ABC and CDA. By A-S-A theorem of congruence of triangles, triangles ABC and CDA are congruent. It follows that |AB| = |CD] and |AD| = |BC|.

2) \Rightarrow 3)

In quadrilateral  ABCD let be AB \| CD and |AB| = |CD|.

Since AC is a traverse of the parallel lines AB and CD, that is \angle{ACD} =\angle{CAB}. The side \overline{AC} is common side of triangles ABC and CDB. By S-A-S theorem of congruence of triangles, triangles ABC and CDB are congruent. It follows that |BC| = |CD|.

 

3) \Rightarrow 4)

In quadrilateral ABCD let be |AB|=|CD| and |BC| = |CD|, and let the point S be the intersection of diagonals \overline{AC} and \overline{BD}.

 

First, consider triangles ABC and CDB. By S-S-S theorem of congruence of triangles, triangles ABC and CDB are congruent. It follows that \angle{ACB} = \angle {CAD}.

Angles ASD and BSC are vertical angles. If now consider triangles ASD and BSD, it follows that \angle{ADS} = \angle{CBS}. Since |BC|= |AD| that triangles ASD and BSC are congruent by A-S-A theorem of congruence triangles. It follows that |AS| = |SC| and |BS| = |SD| which means that point S is the midpoint of \overline{AC} and \overline{BD}.

4) \Rightarrow 5)

In quadrilateral ABCD let the point S be the midpoint of diagonals \overline{AC} and \overline{AD}: |AS|=|CS| and |BS|=|DS|.

Consider triangles BCS and ADS. By S-A-S theorem they are  congruent (|CS| = |AS|\angle{BSC}=\angle{ASB} (vertical angles) – |BS|=|DS|). It follows that \angle{BCS}=\angle{DAS} and \angle{CBS}=\angle{ADS}.

Triangles ABS and CDS are also congruent by S-A-S theorem (|AS|=|CS|\angle{ASB}=\angle{CSD} (vertical angles)  – |BS|=|DS|). It follows that \angle{BAS}=\angle{DCS} and \angle{ABS}=\angle{CDS}.

It follows:

    \[\angle{DAB} = \angle{DAS} + \angle{BAS}= \angle{BCS} + \angle{DCS} = \angle{BCD}\]

and

    \[\angle{ABC} = \angle{ABS} + \angle{CBS}= \angle{CDS} + \angle{ADS} = \angle{ADC}\]

.

5) \Rightarrow 1)

In quadrilateral ABCD let be \alpha=\gamma and \beta = \delta. That means that \alpha+\beta = 180^{\circ} and \gamma+\delta = 180^{\circ}.

Assume that lines AB and AC are not parallel and let the point of intersection of these two lines be the point E which is on the same side of line BC and points to A and D. Then angles \gamma and \delta are interior angles of triangle CDE, but the sum of measures of angles \gamma and \delta is equal to 180^{\circ}, which is a contradiction.

If point E is on the opposite side of line BC and points to A and D, then \alpha + \beta = 180^{\circ}, which is also a contradiction.

It follows that AB \| AC.

Similarly we prove that BC \| AD.

rhombus is a parallelogram which has at least one pair of adjacent sides of equal length.

Opposite angles are of equal measure: \alpha = \gamma, \beta = \delta and that adjacent angles are supplementary.

Diagonals in rhombus are congruent and perpendicular.

 

A kite is a quadrilateral which characterizes two pairs of  sides of equal lengths that are adjacent to each other. Diagonals of a kite are perpendicular and at least one diagonal is a line of symmetry. A kite is also a tangential quadrilateral.

A rectangle is a parallelogram which at least one interior angle is right.

Diagonals in rectangles are congruent.

 

 

 

Square is a rectangle whose all sides are equal.

Diagonals in a square are congruent and perpendicular.

 

Perimeters and areas of  quadrilaterals

Perimeter of any geometric shape is the length of is outline.

Area of any geometric shape is the surface it occupies. Unit of measure for area is m^2 (square meter).

1 square meter is equal to the surface enclosed by a square with sides 1 m.

one square meter

There are also some derived units of measure for areas, for smaller or larger shapes.

1 km^2 = 1 km  \cdot 1 km = 1000 m \cdot 1000 m = 1 000 000 m^2

1 m^2 = 1 m \cdot  1m

1 dm^2 = \frac{1}{10} m^2 \cdot \frac{1}{10} m^2 = \frac{1}{100} m^2

1 cm^2 = \frac{1}{100} m^2 \cdot \frac{1}{100} m^2 = \frac{1}{10 000} m^2

1 mm^2 = \frac{1}{1000} m^2 \cdot  \frac{1}{1000} m^2 = \frac{1}{1 000 000} m^2

 

Area of a square is equal to a square of length of its side.

Area of a rectangular is equal to the product of lengths of adjacent sides.

Area of a rhombus is equal to the product of length of its side and altitude. This is true because, from the picture: if we translate altitude \overline{BE} into point A, and extend side ED over vertex D, we will get triangle E'DA which is congruent with triangle ECB. If we ‘translate’ triangle ECB onto triangle E'DA we will get a rectangular with one side a and other h.

The same that goes for a rhombus works on a parallelogram, the area of a parallelogram is a product of its one side and altitude on that side.

Area of a trapezoid is equal to one half of a product of sum of its bases and altitude.

This formula is a result of dividing a trapezoid into a two triangles AED and BCF, and a rectangle EFCD.

Now, we can write our area as the sum of smaller areas: A_{(ABCD)} = A_{(AED)} + A_{(FBC)} + A_{(EFCD)}.

We know that A_{(EFCD)} =h \cdot c.
Now we need to find A_{(AED)} and A_{(FBC)}. If we translate side b next to AED we get a triangle AHD.

The altitude of a triangle AHD is equal to the altitude of a trapezoid ABCD.

And the side on which this altitude is set is equal to a - c. This leads to a conclusion that:

\ A_ {(AHD)} = h \cdot \frac{a - c}{2}.

This means that:

    \[A_ {(ABCD)} = A_ {(AHD)} + A_ {(HBCD)} =\]

    \[=\frac{h \cdot (a - c)}{2} + c\cdot h= \frac{a \cdot h - h\cdot c + 2h \cdot c}{2} = \frac{h \cdot a + h \cdot c}{2} =\]

    \[=\frac{h \cdot (a + c)}{2}\]

.

 

Quadrilaterals worksheets

  Naming quadrilaterals (278.7 KiB, 416 hits)

  Name the biggest number of quadrilaterals (295.6 KiB, 339 hits)

  Angles in quadrilaterals (423.6 KiB, 333 hits)

  Parallelograms - Find an angle (531.7 KiB, 294 hits)

  Parallelograms - Find an length (547.3 KiB, 286 hits)

  Trapezoids - Find a length of the median (293.1 KiB, 316 hits)

  Trapezoids - Find a length of the half-segment (289.9 KiB, 262 hits)

  Trapezoids - Find a length of a base (312.8 KiB, 295 hits)

  Trapezoids - Angles (284.8 KiB, 399 hits)

  Area of triangles and quadrilaterals (501.9 KiB, 430 hits)

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