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 options 1 and 4.

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: When you have the dimensions of a quadrilateral given along with the diagonal (which is also the hypotenuse) you basically have the quadrilateral divided into two triangles. You can determine if the triangles are going to be right angled or not simply by using the Pythagoras theorem as a test.

Note also that if the angles are right angled, then the quadrilaterals can all be conveniently placed, one in each quarter of the circle without overlapping. This is possible because each quarter of a circle measures 90 degrees.

To test if the angles formed are right angled we shall take the following steps;

Pythagoras theorem states that; AC² = AB² + BC²

Where AC is the longest side (hypotenuse), and AB and BC are the other two sides.

(Quad 1) > 20² = 12² + 15²

400 = 144 + 225

400 ≠ 369

Both sides pf the equation are not equal, hence the vertices of parallelogram 1 do not form a right angle.

(Quad 2)> 34² = 16² + 30²

1156 = 256 + 900

1156 = 1156

Both sides of the equation are equal, hence the vertices of parallelogram 2 forms a right  angle

(Quad 3)> 29² = 20² + 21²

841 = 400 + 441

841 = 841

Both sides of the equation are equal, hence the vertices of parallelogram 3 forms a right angle

(Quad 4)> 26² = 18² + 20²

676 = 324 + 400

676 ≠ 724

Both sides sides of the equation are not equal, hence the vertices of parallelogram 4 do not form a right angle.

From the results shown above, if the four parallelograms are placed in each of the four corners of the circle, they would definitely overlap because two of them do not form right angles at their vertices.

p> q if you calculate p is100 q is 79

answer:

c. the third graph