Quadrilaterals - GMAT Quantitative Reasoning

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.

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.

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.

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.

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:

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.

It is necessary and sufficient to prove that one of the angles of the parallelogram is a right angle.

Assume Statement 1 alone - that . and are supplementary, since they are same-side interior angles of parallel lines. Since , is also supplementary to . But as corresponding angles of parallel lines, . Two angles that are conruent and supplementary are both right angles, so is a right angle.

Assume Statement 2 alone - that and are complementary angles, or, equivalently, . Since the angles of a triangle have measures that add up to , the third angle of , which is , measures , and is a right angle.

Either statement alone proves a right angle and subsequently proves a rectangle.

Assume Statement 1 alone. By congruence, and , making Quadrilateral a parallelogram. However, no clue is given to whether any angles are right or not, so whether the quadrilateral is a rectangle or not remains open.

Assume Statement 2 alone. By congruence, opposite sides , but no clue is provided as to the lengths of opposite sides and . Also, , but no clue is provided as to whether the angles are right. A rectangle would have both characteristics, but so would an isosceles trapezoid with legs and .

Assume both statements are true. Quadrilateral is a parallelogram as a consequence of Statement 1. Since and are consecutive angles of the parallelogram, they are supplementary, but they are also congruent as a consequence of Statement 2. Therefore, they are right angles, and a parallelogram with right angles is a rectangle.

Statement 1 alone is insufficient to answer the question; A quadrilateral in which and are right angles, , and fits the statement, as well as a rectangle, which by defintion has four right angles.

Statement 2 alone is insufficient as well, as a parallelogram with acute and obtuse angles, as well as a rectangle, fits the description.

Assume both statements, and construct diagonal to form two triangles and . By Statement 1, both triangles are right with congruent legs , and congruent hypotenuses, both being the same segment . By the Hypotenuse Leg Theorem, . By congruence, . The quadrilateral, having two sets of congruent opposite sides, is a parallelogram; a parallelogram with right angles is a rectangle.

When a square and rectangle have the same perimeter, the square will have a larger area because having 4 equal sides maximizes the area. However, from statement 1, it is impossible to tell if the rectangle is also a square. When the information from statement 2 is combined, we can conclude that the rectangle is not also a square.

To find the rectangle, you need the length and the width.

If you know the diagonal and the angle it forms with one of the longer sides, you can use trigonometry to find both length and width:

From there, the area follows.

If you know only the diagonal, you have insufficient information; the length and width can vary according to that angle. If you only know the angle, you can discern the proportions of the sides, but not the actual lengths.

The portion of the rectangle to the right of the -axis has area ; to the left, . From Statement 1 alone, since , , and the portion of the rectangle on the right is greater than the portion on the left. However, this is all we can determine.

By a similar argument, from Statement 2 alone, the portion of the rectangle above is greater than the portion below, but this is all we can determine.

From both statements together, however, we can compare the portions of the rectangles in the four quadrants. The areas of each are:

Quadrant I:

Quadrant II:

Quadrant III:

Quadrant IV:

Since and , is the greatest of the four quantities, and we see that Quadrant I includes the lion's share of the rectangle.

The figure referenced is below:

Assume Statement 1 alone. is a diagonal of the rectangle, and also a diameter of the circle. Statement 1 gives the area of the circle, from which the radius, and, subsequently, , can be calculated. However, infinitely many rectangles of different areas can be constructed in this circle, so without any further information, it is not clear what the sidelengths are - and what the area is.

Assume Statement 2 alone. This statement only gives the length of one side. Without any further information, the area of the rectangle is unknown.

Now assume both statements are true. can be calculated, and is given, so the Pythagorean Theorem can be used to find . The area of the rectangle is the product , so the two statements together are sufficient.

The figure referenced is below (note that the figure itself assumes Statement 2, but this is not known from Statement 1):

Assume Statement 1 alone. A diagonal of a rectangle inscribed inside a circle is a diameter of the circle. Statement 1 gives the area of the circle, from which the radius, and, subsequently, , can be calculated. However, infinitely many rectangles of different areas can be constructed in a given circle, so without any further information, it is not clear what the sidelengths are - and what the area is.

Assume Statement 2 alone. It follows that all of the sides of the rectangle/square are congruent, but without the common sidelength, the area of the square cannot be calculated.

Assume both statements. can be calculated, and the area of the square can be calculated to be .

A a rectangle with sides of length 1 and 4 and a rectangle with sides of length 2 and 3 both have perimeter 10, but they have different areas ( 4 and 6, respectively), making Statement 1 alone inconclusive. Statement 2 is inconclusive, there being infinitely many primes.

Assume both statements.

Then

Since and are both prime integers, one must be 2 and the other must be 3 (1 and 4 cannot be a possibility, since 1 is not a prime). It does not matter which is which, so the numbers can be multiplied to obtain area 6.

Assume both statements. Then from Statement 1, it follows that:

There are two pairs of odd primes that add up to 18 - (5,13), in which case the area is 65, and (7,11), in which case the area is 77. The two statements together are inconclusive.

Because all 4 sides of a square are equal, knowing the length of one side is sufficient to answer the question. Using the Pythagorean Theorem, you can calculate the length of 1 side of a square by knowing the length of a diagonal and then calculate the area.

The area of the black region is one-fourth the difference of the areas of the circle and the square.

If is the sidelength of the square, then the length of its diagonal - which is also the diameter of the circle - is, by the Pythagorean Theorem, , and the radius . Therefore, if you calculate either the radius or the sidelength, you can calculate the other, allowing you to find the areas of the circle and the square.

Statement 1 allows you to find the sidelength; just divide 40 by 4.

Statement 2 allows you to find the radius; just solve for in the equation .

Therefore, either one gives you enough information to solve the problem.

The diameter of the inscribed circle is equal to the the sidelength of the square. From Statement 1, the circumference can be divided by to obtain this measure, and this can be squared to obtain the area of the square.

Statement 2 gives extraneous information, as, by regularity of the figure, it is already known that is one fourth of the circle and, subsequently, a arc.