Question

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

Answer 1

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

Answer 2

In this exercise we have to use the knowledge of parallelogram to identify which equations are not:

Letter B and D

Given the following parallelograms we have:

A)$P1: 12^2+15^2=20^2 \\144+225=400 \\369=400$

B) $P2: 16^2+30^2=34^2\\256+900=1156 \\ 1156=1156$

C)$P3: 20^2+21^2=29^2 \\ 400+441=841 \\841=841$

D) $P4: 18^2+20^2=26^2 \\ 324+400=676 \\724=676$

So we have to:

• 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 4 do not form right angles.

See more about parallelogram at brainly.com/question/1563728