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tensa zangetsu [6.8K]
4 years ago
11

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. mr025-1.jpg 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
2 answers:
mr_godi [17]4 years ago
8 0
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
Alchen [17]4 years ago
4 0

Answer:

B. 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.

E. 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.

Step-by-step explanation:

P1: 12^2+15^2=20^2  144+225=400  369=400

P2: 16^2+30^2=34^2  256+900=1156  1156=1156

P3: 20^2+21^2=29^2  400+441=841  841=841

P4: 18^2+20^2=26^2  324+400=676  724=676

So the answers are B and E

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Consider the contrapositive of the statement you want to prove.

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In this case, the contrapositive claims that

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