GT Geometry Unit 6: Quadrilaterals Jeopardy. Angles of Polygons || - ogram Properties || - ogram...
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Transcript of GT Geometry Unit 6: Quadrilaterals Jeopardy. Angles of Polygons || - ogram Properties || - ogram...
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GT Geometry
Unit 6: Quadrilaterals
Jeopardy
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Angles of Polygons
|| - ogram Properties
|| - ogram Tests
Rhombi/Trapezoids
AreaCoordinate
Plane
100 100 100 100 100 100
200 200 200 200 200 200
300 300 300 300 300 300
400 400 400 400 400 400
500 500 500 500 500 500
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$100
What is the sum of the interior angles of a pentagon?
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$100
540 Degrees
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$200
What is the measure of each exterior angle of a regular octagon?
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$200
45 degrees
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$300
If each interior angle of a regular polygon is 140 degrees how many sides does the polygon have?
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$300
9 sides
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$400
If each exterior angle of a regular polygon is 72 degrees how many sides does the polygon have?
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$400
5 sides
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$500
If each interior angle of a regular polygon is 150 degrees what is the measure of each exterior angle?
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$500
30 degrees
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$100
Find x if the quad below is a parallelogram
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$100
X = 7
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$200
Find x if the quad below is a parallelogram
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$200
X = 12
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$300
Find x if the quad below is a parallelogram
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$300
X = 3
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$400
Find x if the quadrilateral below is a parallelogram
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$400
X= 33
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$500Find x if the quadrilateral below is a
parallelogram
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$500
X = 14.5
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$100
Can we prove this quadrilateral is a parallelogram?
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$100
Yes both pairs of opposite sides are congruent
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$200
Can we prove this quadrilateral is a parallelogram?
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$200
No, we don’t know that both pairs of opposite angles are congruent
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$300
Can we prove this quadrilateral is a parallelogram?
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$300
Yes, one pair of opposites sides is both congruent and parallel
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$400
Can we prove this quadrilateral is a parallelogram?
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$400
Yes, diagonals bisect each other
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$500
Can we prove this quadrilateral is a parallelogram?
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$500
Yes, the total sum of the angles of a quadrilateral is 360 degrees. Therefore x = 100. Since the opposite angles are congruent it is a parallelogram
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$100
If the quadrilateral below is a rhombus, find x
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$100
X = 4.5
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$200
If the trapezoid below is an isoceles trapezoid find x.
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$200
X = 12
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$300
If the trapezoid below is an isosceles trapezoid, find x
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$300
X = 14.5
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$400
If the quadrilateral below is a rhombus find x
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$400
X = 2
Diagonals of a rhombus bisect angles
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$500
If the quadrilateral below is a rhombus find x
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$500
X = 17
The diagonals of a rhombus are perpendicular so use the
Pythagorean theorem
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$100
Find the area of the polygon
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$100
A = 70Area of a rhombus = ½ (d1)(d2)
D1 = 7 + 7 = 14
D2 = 5+5 = 10
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$200
Find the area of the quadrilateral
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$200
A = 64 Area of a trapezoid = ½ (b1+b2)h
= ½ ( 6+10) 8
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$300
Find the area of the quadrilateral below. Hint (use the Pythagorean theorem to find the missing side.)
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$300
A = 192The height of the rectangle = 12.
12 x 16 = 192
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$400
The quadrilateral has an area of 60 sq inches. Find x
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$400
x = 8
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$500
Find the area of the yellow region.
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X = 96.
The area of the rectangle = 16 x 12. The area of the two triangles are ½ (8)(12). Subtract the
two.
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$100
JKLM is a quadrilateral with
J(0,0), K (3,7), L(9,7) and M(6,0).
Is JKLM a parallelogram?
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$100
Yes opposite sides are parallel and congruent
Slope: JK = 7/3LM = 7/3KL = 0 JM = 0
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$200
Is ABCD a rhombus?
A (3,1) B(3,-3) C(-2,-3) D (-2,1)
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$200
No. Diagonals are not perpendicular
Slope of diagonals
AC = 4/5
BD = - 4/5
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$300
Is LMNO a trapezoid?
L ( 5,2) M (1,9) N (-3, 2) O (1,-5)
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$300
Yes. 1 opposite side is parallel. It is also an isosceles trapezoid.
Slope Congruent Legs
LM = - 7/4 LO = sq rt 65
ON = -7/4 MN = sq rt 65
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$400
Is PQRS a square?
P (5,2) Q (2,5) R( -1,2) S (2,-1)
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Yes. Diagonals are congruent and perpendicular
Congruent Slope
RP = 6 RP = 0
QS = 6 QS = undefined.
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$500
JKLM is a quadrilateral with
K(6,0) L (7,2) and M (2,8) what are the coordinates of J to make JKLM a parallelogram?
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$500
J = (1,6) or (11,-6)The slope of LM = -6/5. Therefore the slope of JK = -6/5. The slope could
also be written as 6/-5. Therefore we must solve for x and y. The following two coordinates would make this slope (1,6) or (11,-6)
y – 0 = 6 x = 1, y = 6 y – 0 = -6 x = 11, y = -6
x – 6 = -5 x – 6 = 5
Then we find that the distance for LM = sq rt 61. Therefore, we plug in both possible coordinates to determine which one gives us a distance for JK = sq rt 61. Since they both do both answers are correct.
JK when J = (1,6) = sq rt 61 JK when J = 11,-6) = sq rt 61