DSQ: Calculating an angle in a quadrilateral - GMAT Quantitative

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.

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.

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.