Two opposite angles of a quadrilateral inscribed inside a circle are supplementary, so 

The four sides of a rhombus are congruent, so a diagonal of the rhombus cuts it into two isosceles triangles. The two angles adjacent to the diagonal are congruent, so the third angle, the one marked, measures:

To prove that Polygon  is also a rectangle, we need to prove that any one of its angles is a right angle.

If , then by definition of perpendicular lines,  is right.

If , then, since  and  form a linear pair,  is right.

If , then, by the Converse of the Pythagorean Theorem,  is a right triangle with right angle .

If  and  are complementary angles, then, since 

, making  right.

However, since, by definition of a parallelogram, , by the Alternate Interior Angles Theorem,  regardless of whether the parallelogram is a rectangle or not.

Case 1: The two angles are opposite angles of the parallelogram. In this case, they are congruent, and

Case 2: The two angles are consecutive angles of the parallelogram. In this case, they are supplementary, and

By the Pythagorean Theorem, 

Also by the Pythagorean Theorem, 

The perimeter of Quadrilateral  is

A rhombus has four congruent sides. Statement 1 gives the length of one of them, so that length can be multiplied by  to yield the perimeter.

The area of a rhombus alone has no bearing on its perimeter, so Statement 2 alone is insufficient.

In an isosceles trapezoid, the two non-parallel sides are equal. In this case, they are both  long. To find the perimeter, we also need to know the length of both bases. We are told that Base B is . Eliminate any options less than or equal to .

To find the length of Base A, we need to translate the following: "The length of one of its bases (Base A) is equal to three times the amount of seven inches fewer than the length of the other base (Base B)." For translating, break the statement down:

Starting with ,

"seven inches fewer than the length of the other base (Base B)": 

"three times the amount of seven inches fewer than the length of the other base (Base B)": 

So, Base A is  long, so the perimeter of isosceles trapezoid  is:

This is a perimeter problem, but first we need to find our side lengths. 

The short sides are  meters. The long sides are two times 4 more than the short sides. So

"four more" 

"two times" meters

So our two side lengths are 120 and 56 meters. Find the perimeter by the following:

So, 352 meters

The rhombus is a special kind of a parallelogram. Its sides are all of the same length. Therefore, we just need to find one length of this quadrilateral. To do so, we can apply the Pythagorean Theorem on triangle AEC for example, since we know the length of the diagonals. Also, the diagonals intersect at their center. Therefore, triangle AEC has length,  and . Therefore,  or . The perimeter is then .

For a triangle to be a right triangle, the sides must obey the Pythagorean Theorem. Let's try our options.

3, 4, 5: You should know this is a right triangle without having to do any calculations because it is one of the special triangles that you should remember. But if you didn't, .

28, 45, 53: 

45, 55, 75:  . The sides don't follow the Pythagorean Theorem so this can't be a right triangle. This is our answer. Let's check the remaining two sets of sides as well.

48, 64, 80:  . These are pretty big numbers and this math might take a while. Instead of doing these calculations, we could also see if 48, 64, and 80 look like any of the special triangles we know. Let's divide the three numbers by 16. 48/16 = 3, 64/16 = 4, and 80/16 = 5. Then this is just a type of 3,4,5 triangle, which we know is a right triangle.

84, 35, 91:  . Again, these are big numbers to square. Let's divide the three numbers by their greatest common factor, 7. 84/7 = 12, 35/7 = 5, 91/7 = 13. Then this is a 5, 12, 13 triangle, which is another of our special triangles that we know is a right triangle.