To find the perimeter of the quadrilateral, we need to know whether it is of a special type of quadrilaterals and we need to know the length of the sides.

Statement 1 tells us only that the quadrilateral is a rhombus. Indeed, a quadrilateral with perpendicular diagonals intersecting at their midpoint must be a rhombus. However we don't know any length of the sides. 

Statement 2 says gives us the length us two consecutive sides. It could be tempting to answer that it is sufficient, however, we can't conclude that the quadrilateral has equal lengths. Therefore this statement alone is insufficient.

Both statements together are sufficient since we can conclude that the quadrilateral is a rhombus, and twice   will give us the perimeter.

To find perimeter, we need to find the length of all the sides. Recall that rectangles are made up of two pairs of equal sides.

I) Relates one side to another non-equivalent side.

II) Gives us side , which must be equivalent to .

Use II) and I) to find all the side lengths, then add them up. Both are needed.

Recap:

Consider rectangle CONT

I) Side CO is three fourths of side ON

II) Side NT is 15.7 meters long

What is the perimeter of CONT? 

Because we are dealing with a rectangle, we know the following:

Find perimeter with:

Use I) and II) to write the following equation:

So:

And finally:

Statement 1): The area of the rectangle is 24.

Write the area for a rectangle and substitute the value of the area.

The length and width of the rectangle are unknown, and each set of dimensions will provide a different perimeter.  This statement is insufficient to find the perimeter of the rectangle.

Statement 2): The diagonal of the rectangle is 5.

Given the diagonal of the rectangle, the Pythagorean Theorem can be used to solve for the diagonal.  Express the equation in terms of length and width.

Similar to the case in Statement 1), both the length and width are unknown, and the equation by itself is insufficient to solve for the perimeter of the rectangle.

Attempting to use both equations:  and  to solve for length and width will yield complex numbers as part of the solution.

Therefore:

Any two consecutive angles of a parallelogram are supplementary, so if one angle has measure , all angles can be proven to have measure . This is the definition of a rectangle. Statement 1 therefore proves the parallelogram to be a rectangle.

Also, a parallelogram is a rectangle if and only if its diagonals are congruent, which is what Statement 2 asserts. 

From either statement, it follows that parallelogram  is a rectangle.

A circle can be circumscribed about a quadrilateral if and only if both pairs of opposite angles are supplementary. This is not proved or disproved by Statement 1 alone:

Case 1:

This is not a rectangle, and opposite angles are supplementary, so a circle can be constructed to circumscribe the quadrilateral.

Case 2:   

This is not a rectangle, and opposite angles are not supplementary, so a circle cannot be constructed to circumscribe the quadrilateral. 

From Statement 2, however, it follows that a two opposite angles are not a supplementary pair, so a circle cannot be circumscribed about it.

For the diagonals of a quadrilateral to be perpendicular, the quadrilateral must be a kite or a rhombus - in either case, there must be two pairs of adjacent congruent sides. Neither statement alone proves this, but both statements together do.

 and , the diagonals of Parallelogram  , are perpendicular if and only if Parallelogram  is also a rhombus.

Opposite sides of a parallelogram are congruent, so if Statement 1 is assumed, . Parallelogram  a rhombus; subsequently,  . 

The angle measures are irrelevant, so Statement 2 is unhelpful.

From Statement 1 alone, we can calculate ,  since two opposite angles of a quadrilateral inscribed inside a circle are supplementary:

From Statement 2 alone, we can calculate , since the degree measure of an inscribed angle of a circle is half that of the arc it intersects:

A circle can be circumscribed about a quadrilateral if and only if both pairs of its opposite angles are supplementary.

An isosceles trapezoid has this characteristic. Assume without loss of generality that  and  are the pairs of base angles. 

Then, since base angles are congruent,  and . Since the bases of a trapezoid are parallel, from the Same-Side Interior Angles Theorem,  and  are supplementary, and, subsequently, so are   and , as well as  and  .

If , then  and  form a supplementary pair, as their measures total ; since the measures of the angles of a quadrilateral total , the measures of   and  also total , making them supplementary as well.

Therefore, it follows from either statement that both pairs of opposite angles are supplementary, and that a circle can be circumscribed about the quadrilateral.

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

If we assume Statement 1 alone, that , then, since  and  form a linear pair,  is right. 

If we assume Statement 2 alone, that , it follows from the converse of the Pythagorean Theorem that  is a right triangle with right angle .

Either way, we have proved that the parallelogram is a rectangle.