GT Geometry Unit 6: Quadrilaterals Jeopardy

GT Geometry

Unit 6: Quadrilaterals

Jeopardy

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

$100

What is the sum of the interior angles of a pentagon?

$100

540 Degrees

$200

What is the measure of each exterior angle of a regular octagon?

$200

45 degrees

$300

If each interior angle of a regular polygon is 140 degrees how many sides does the polygon have?

$300

9 sides

$400

If each exterior angle of a regular polygon is 72 degrees how many sides does the polygon have?

$400

5 sides

$500

If each interior angle of a regular polygon is 150 degrees what is the measure of each exterior angle?

$500

30 degrees

$100

Find x if the quad below is a parallelogram

$100

X = 7

$200

Find x if the quad below is a parallelogram

$200

X = 12

$300

Find x if the quad below is a parallelogram

$300

X = 3

$400

Find x if the quadrilateral below is a parallelogram

$400

X= 33

$500Find x if the quadrilateral below is a

parallelogram

$500

X = 14.5

$100

Can we prove this quadrilateral is a parallelogram?

$100

Yes both pairs of opposite sides are congruent

$200

Can we prove this quadrilateral is a parallelogram?

$200

No, we don’t know that both pairs of opposite angles are congruent

$300

Can we prove this quadrilateral is a parallelogram?

$300

Yes, one pair of opposites sides is both congruent and parallel

$400

Can we prove this quadrilateral is a parallelogram?

$400

Yes, diagonals bisect each other

$500

Can we prove this quadrilateral is a parallelogram?

$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

$100

If the quadrilateral below is a rhombus, find x

$100

X = 4.5

$200

If the trapezoid below is an isoceles trapezoid find x.

$200

X = 12

$300

If the trapezoid below is an isosceles trapezoid, find x

$300

X = 14.5

$400

If the quadrilateral below is a rhombus find x

$400

X = 2

Diagonals of a rhombus bisect angles

$500

If the quadrilateral below is a rhombus find x

$500

X = 17

The diagonals of a rhombus are perpendicular so use the

Pythagorean theorem

$100

Find the area of the polygon

$100

A = 70Area of a rhombus = ½ (d1)(d2)

D1 = 7 + 7 = 14

D2 = 5+5 = 10

$200

Find the area of the quadrilateral

$200

A = 64 Area of a trapezoid = ½ (b1+b2)h

= ½ ( 6+10) 8

$300

Find the area of the quadrilateral below. Hint (use the Pythagorean theorem to find the missing side.)

$300

A = 192The height of the rectangle = 12.

12 x 16 = 192

$400

The quadrilateral has an area of 60 sq inches. Find x

$400

x = 8

$500

Find the area of the yellow region.

$500

X = 96.

The area of the rectangle = 16 x 12. The area of the two triangles are ½ (8)(12). Subtract the

two.

$100

JKLM is a quadrilateral with

J(0,0), K (3,7), L(9,7) and M(6,0).

Is JKLM a parallelogram?

$100

Yes opposite sides are parallel and congruent

Slope: JK = 7/3LM = 7/3KL = 0 JM = 0

$200

Is ABCD a rhombus?

A (3,1) B(3,-3) C(-2,-3) D (-2,1)

$200

No. Diagonals are not perpendicular

Slope of diagonals

AC = 4/5

BD = - 4/5

$300

Is LMNO a trapezoid?

L ( 5,2) M (1,9) N (-3, 2) O (1,-5)

$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

$400

Is PQRS a square?

P (5,2) Q (2,5) R( -1,2) S (2,-1)

$400

Yes. Diagonals are congruent and perpendicular

Congruent Slope

RP = 6 RP = 0

QS = 6 QS = undefined.

$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?

$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