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. 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.

The correct options are;

*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.

Step-by-step explanation: What James is trying to do is quite simple which is, he wants to place four quadrilaterals inside a circle and he wants the vertices to touch one another at the center of the circle without having to overlap.

This is possible and quite simple, provided all the quadrilaterals have right angles (90 degrees). This is because the center of the circle measures 360 degrees and we can only have four vertices placed there without overlapping only if they all measure 90 degrees each (that is, 90 times 4 equals 360).

We can now show whether or not all four parallelograms have right angles by applying the Pythagoras' theorem to each of them. Note that James has cut the shapes in such a way that the hypotenuse (diagonal) and the other two legs have already been given in the question. As a reminder, the  Pythagoras' theorem is given as,

AC² = AB² + BC² Where AC is the hypotenuse (diagonal) and AB and BC are the other two legs. The experiment would now be as follows;

Quadrilateral 1;

20² = 12² + 15²

400 = 144 + 225

400 ≠ 369

Therefore the vertices of parallelogram 1 do not form a right angle

Quadrilateral 2;

34² = 16² + 30²

1156 = 256 + 900

1156 = 1156

Therefore the vertices of parallelogram 2 forms a right angle

Quadrilateral 3;

29² = 20² + 21²

841 = 400 + 441

841 = 841

Therefore the vertices of parallelogram 3 forms a right angle

Quadrilateral 4;

26² = 18² + 20²

676 = 324 + 400

676 ≠ 724

Therefore the vertices of parallelogram 4 do not form a right angle

The results above shows that only two of the parallelograms cut out have right angles (like a proper square or rectangle for instance), while the other two do not have right angles.

Therefore, the correct option are as follows;

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.

B and E

Step-by-step explanation:

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.

what is the question