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Zarrin [17]
3 years ago
13

James cut out four parallelograms, the dimensions of which are shown below. Parallelogram 1 length: 12 in. width: 15 in. diagona

l: 20 in. Parallelogram 2 length: 16 in. width: 30 in. diagonal: 34 in. Parallelogram 3 length: 20 in. width: 21 in. diagonal: 29 in. Parallelogram 4 length: 18 in. width: 20 in. diagonal: 26 in. James put the parallelograms together so one vertex from each paper exists on a point, as shown in the circle. 4 parallelograms are put together so that one vertex from each paper exists on a point. Which statement explains whether or not the parallelgrams can be put together so each occupies one-quarter of the area of the circle without overlapping any other pieces? Check all that apply. The quadrilaterals can be placed such that each occupies one-quarter of the circle. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 1 do not form right angles. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 2 do not form right angles. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 3 do not form right angles. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 4 do not form right angles.

Mathematics
1 answer:
Alenkinab [10]3 years ago
3 0

Answer: by the looks of this its 2

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HJ
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Answer:

19 degrees

Step-by-step explanation:

From the question given

The interior angles are x+12 and x - 3

Exterior angle is <IJK = 5x-6

Using the rule that states that the sum of interior angle of a triangle is equal to the exterior

<JHI + <HIJ = <IJK

x+12 + x-3 = 5x - 6

2x+9 = 5x -6

2x - 5x = -6-9

-3x = -15

x = -15/-3

x = 5

Get <IJK

Recall that <IJK = 5x - 6

<IJK = 5(5) - 6

<IJK = 25-6

<IJK = 19 degrees

Hence the measure of <IJK is 19 degrees

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