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Quadrilateral Handbook

Geometry Hour 4
by

Megan Martin

on 3 April 2014

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Transcript of Quadrilateral Handbook

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)
Full transcript