**Quadrilateral Handbook**

**Quadrilaterals**

**These are an essential to not only geometry, but life. So what is it? A quadrilateral is a figure with four sides, simple right. Well as we discover specialized quadrilaterals, we realize they are more complex than we give credit.**

**Parallelograms**

Properties:

Sides

opposite sides are congruent

opposite pairs of sides are parallel

Diagonals

diagonals bisect each other

Angles

opposite angles are congruent

consecutive angles are supplementary

Rectangles

Properties:

Sides

opposite sides are congruent

opposite sides are parallel

Diagonals

diagonals bisect each other and are congruent

Angles

all angles are congruent

all angles are 90 degrees

all angles are supplementary

Trapezoids

Properties:

sides

one pair of parallel sides

Kites

Properties:

sides

Two congruent pairs of sides

diagonals

perpendicular diagonals

angles

opposite angles that are not the ends are congruent

Some quadrilaterals are more specific than others, thus the quadrilateral hierchy was born. This is a tool to help us know the properties of each. For example, a rhombus not only has two parallel sides like a parallelogram, but two congruent sides, like a kite. Why? Because it is higher in the hierchy.

Prove:

opposite sides are congruent

Prove:

opposite sides are parallel

Prove:

Diagonals Bisect each other and are congruent

Prove:

One pair of parallel sides

Isosceles Trapezoids

Properties:

sides

one pair of parallel sides

one pair of opposite sides are congruent

diagonals

diagonals are congruent

angles

base angles are congruen

t

Prove:

Two congruent pairs of sides

Prove:

Perpendicular diagonals

Prove:

one pair of opposite sides are congruent

Prove:

diagonals are congruent

Rhombuses

Properties:

sides

All sides are congruent

diagonals

diagonals are perpendicular bisectors

angles

opposite angles that are congruent

adjacent angles are supplementary

Prove:

all sides are congruent

Prove:

The diagonals are perpendicular bisectors

Squares

Properties:

sides

all sides are congruent

opposite sides are parallel

diagonals

they are perpendicular bisectors

are congruent

angles

all angles are 90 degrees/right

all angles are supplementary

Prove:

all sides are congruent

Prove:

opposite sides are parallel

Prove:

diagonals are perpendicular bisectors

BY Megan Martin

Prove:

the diagonals are congruent

See Parallelogram

See Parallelogram

For bisect see parallelogram

For congruent see isosceles trapezoid

For perpendicular see kite

For bisectors see parallelogram

See Rhombus

See Parallelogram

See Rhombus

See Isosceles Trapeziod

See Parallelogram for how to solve

See parallelogram for how to solve

See parallelogram for how to solve

See Parallelogram for how to solve

Applying Properties of Parallelograms:

A

B

C

D

1.) Say m<DAB=40, you can assume this adds up to 180 because by the properties of a //agram, you know it is a SSIA, leading to the conclusion m<ADC=140. The argument is the same for the other angles

2.) If AD=4cm and DC=6cm then you know, because of the properties of a //agram, that BC=AD and AB=DC

3.) Imagine that there is a point E at the intersection of diagonals AC and BD. If AE=2cm and DE=6cm, then, by the properties of //agrams, EC=2cm and EB=6cm

Applying Properties of Rectangles:

Prove:

Opposite sides are congruent

Prove:

Opposite pairs of parallel sides

Prove:

The diagonals bisect each other

A

B

C

D

1.) If given that m<A=90, you would hopefully know that all the other angles are 90 by the properties of rectangles

2.) If AD=5cm and DC=8cm, then by the properties of rectangles AD=BC and DC=AB, so BC=5cm AB=8cm

3.) Imagine point E is at the intersection of diagonals AC and BD, AE= 10cm, by the properties of a rectangle, you know that the diagonals are congruent and bisect eachother, therefore all the diagonal sections =10cm

Applying the properties of Isosceles Trapezoids:

1.) If m<ADC=60, you know that AB // to DB by the definition of trapeziods, so by SSIA m<DAB=120, the argument is the same for the remaining angles

A

B

C

D

2.)If AD=3cm, then BC=3cm because of the properties of isosceles trapezoids

3) If diagonal AC is 9cm, then DB is also 9cm because of the properties of isosceles trapezoids, the diagonals being congruent

Applying properties of Kites:

A

B

C

D

1.) If m<ADC=90 , then m<ABC=90 because of the Kite Diagonal Theorem, and also because of this and the given (m<DAC=30) that m<DAB=60 by adding the two angles together. And from the triangle sum theorem, m<DCA=60, and that m<DCB=120 by the angle addition postulate.

2.) If DC is 25m and AD=2m, then by the kite symmetry theorem, BC=25m and AB=2m

3.) If point E is the intersection of the diagonals, then DE=9km, then EB=9km by the kite symmetry theorem.

Applying the properties of rhombuses:

1.) Consider Rhombus ABCD, m<DAE=33, you know that m<AED=90 by the properties of a rhombus (the diagonals are perpendicular bisectors). By the triangle sum theorem we know m<EDA=57. The argument is the same for the other triangles.

A

B

C

D

E

2.) If AD=60in, then AD=DC=CB=BA=60in by the definition of rhombus

3.) If AE=8cm, then EC also =8cm because the diagonals bisect each other in a rhombus.

Applying the properties of a square:

1.) If m<A=90 you know that all the angles are 90 because that is a property of a square

A

B

C

D

2.) If AD=5cm then all of the sides of the square are 5cm because that is another property of squares

3. Point E is the intersection of the diagonals AC and DB.

If AE=70cm, then all the other segments that are included in the diagonals and are bisected by E are congruent. This is because the diagonals of a square are perpendicular bisectors and congruent.

E

A

B

C

D

A= (2,2)

B= (6,2)

C= (4,0)

D= (0,0)

Use the Distance formula to prove this

NOTE: /numbers\ = square root

'2= square

AB=

/(2-6)'2+(2-2)'2\

/-4'2+0\

/16\

4

DC=

/(0-4)'2+(0-0)'2\

/-4'2+0)

/16\

4

Same argument for other side

Use parallelogram ABCD (seen in previous problem)

A= (2,2)

B= (6,2)

C= (4,0)

D= (0,0)

Slope=Y2 -Y1/ X2-X1

AC=

(2-2)/(6-2)

0/4

Slope=0

DC=

(0-0)/(0-4)

0/-4

Slope=0

A= (2,2)

B= (6,2)

C= (4,0)

D= (0,0)

We are going to keep Parallelogram ABCD for this as well

First find the distance of AC:

/(2-4)'2+(2-0)'2\

/-2'2+2'2\

/4+4\

/8\

which it is about 3

same argument for other pair

Now check your parallelogram to see if this matches up

A

B

C

D

A= (0,2)

B= (4,2)

C= (4,0)

D= (0,0)

Use the DISTANCE FORMULA

From trapezoid ABCD (see the applying properties practice problems) with coordinates:

A= (-1,2)

B= (1,2)

C= (2,0)

D= (-2,0)

AC=

/(-1-2)'2+(2-0)'2\

/-3'2+2'2\

/9+4\

/13\

=3.6

BD=

/(1+2)'2+(2-0)'2

/3'2+2'2\

/9+4\

/13\

=3.6

Using the kite that is on the Applying Properties, find the slope of the diagonals

A= (2,6)

B= (4,4)

C= (2,1)

D= (0,4)

AC=

(1-6)/(2-2)

-5/0

BC=

(4-4)/(0-4)

0/4

The slopes are opposite reciprecals, meaning they are perpendicular

A= (0,6)

B= (5,0)

C= (0,-6)

D= (-5,0)