DSQ: Calculating an angle in a quadrilateral - GMAT Quantitative

Card 0 of 80

Is parallelogram a rectangle?

Statement 1: $\angle A = $90^{\circ }$$

Statement 2:

Tap to see back →

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.

Given a quadrilateral , can a circle be circumscribed about it?

Statement 1: Quadrilateral is not a rectangle.

Statement 2: $m\angle Q = $120^{\circ }$, m \angle A = $50^{\circ }$$

Tap to see back →

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: $m\angle Q = $120^{\circ }$, m\angle U = $120^{\circ }$, m \angle A = $60^{\circ }$, m \angle D = $60^{\circ }$$

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

Case 2: $m\angle Q = $120^{\circ }$, m\angle U = $110^{\circ }$, m \angle A = $70^{\circ }$, m \angle D = $60^{\circ }$$

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.

Given a quadrilateral , can a circle be circumscribed about it?

Statement 1: Quadrilateral is an isosceles trapezoid.

Statement 2: $m\angle Q = $120^{\circ }$, m \angle A = $60^{\circ }$$

Tap to see back →

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 $\angle Q, \angle U$ and $\angle A, \angle D$ are the pairs of base angles.

Then, since base angles are congruent, $m\angle Q =m \angle U$ and $m\angle A =m \angle D$ . Since the bases of a trapezoid are parallel, from the Same-Side Interior Angles Theorem, $\angle Q$ and $\angle D$ are supplementary, and, subsequently, so are $\angle Q$ and $\angle A$ , as well as $\angle U$ and $\angle D$ .

If $m\angle Q = $120^{\circ }$, m \angle A = $60^{\circ }$$ , then $\angle Q$ and $\angle A$ form a supplementary pair, as their measures total ; since the measures of the angles of a quadrilateral total , the measures of $\angle U$ and $\angle D$ 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.

Given Parallelogram .

True or false: $\overline{AC}$ \perp $\overline{BD}$

Statement 1:

Statement 2: $m \angle ABC = $45^{\circ }$$

Tap to see back →

$\overline{AC}$ and $\overline{BD}$ , 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, $\overline{AC}$ \perp $\overline{BD}$ .

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

Are the diagonals of Quadrilateral perpendicular?

(a)

(b)

Tap to see back →

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.

Quadrilateral is inscribed in a circle.

What is $m \angle Q$ ?

Statement 1: $m \angle A = 140 ^{\circ }$

Statement 2: $m \widehat{UAD} = 80 ^{\circ }$

Tap to see back →

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

$m \angle Q = $180^{\circ }$ - m \angle A = $180^{\circ }$ - $140^{\circ }$ = $40^{\circ }$$

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

$m \angle Q = $\frac{1}{2}$\cdot m \widehat{UAD} = $\frac{1}{2}$\cdot 80 ^{\circ } = 40 ^{\circ }$

The above shows Parallelogram . Is it a rectangle?

Statement 1: $m \angle 1 = 90 ^{\circ }$

Statement 2:

Tap to see back →

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 $m \angle 1 = $90^{\circ }$$ , then, since $\angle 1$ and $\angle BCD$ form a linear pair, $\angle BCD$ is right.

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

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

Refer to the above figure. You are given that Polygon is a parallelogram but not that it is a rectangle. Is it a rectangle?

Statement 1: $\angle BAD \cong \angle 1$

Statement 2: $\angle DAC$ and $\angle DCA$ are complementary angles.

Tap to see back →

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

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

Assume Statement 2 alone - that $\angle DAC$ and $\angle DCA$ are complementary angles, or, equivalently, $m \angle DAC + m \angle DCA = 90 ^{\circ }$ . Since the angles of a triangle have measures that add up to , the third angle of $\Delta DAC$ , which is $\angle D$ , measures $90 ^{\circ }$ , and is a right angle.

Either statement alone proves $\angle D$ a right angle and subsequently proves a rectangle.

True or false: Quadrilateral is a rectangle.

Statement 1: $\bigtriangleup QUA \cong \bigtriangleup ADQ$

Statement 2: $\bigtriangleup QUA \cong \bigtriangleup UQD$

Tap to see back →

Assume Statement 1 alone. By congruence, $\overline{QU}$ \cong $\overline{AD}$ and $\overline{QD}$ \cong $\overline{AU}$ , 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 $\overline{QD}$ \cong $\overline{UA}$ , but no clue is provided as to the lengths of opposite sides $\overline{Q U}$ and $\overline{DA}$ . Also, $\angle QUA \cong \angle UQD$ , 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 $\overline{Q U}$ and $\overline{DA}$ .

Assume both statements are true. Quadrilateral is a parallelogram as a consequence of Statement 1. Since $\angle QUA$ and $\angle UQD$ 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.

True or false: Quadrilateral is a rectangle.

Statement 1: $\angle Q$ and $\angle A$ are right angles.

Statement 2: $\overline{QU}$ \cong $\overline{AD}$

Tap to see back →

Statement 1 alone is insufficient to answer the question; A quadrilateral in which $\angle Q$ and $\angle A$ are right angles, $m \angle U = $91^{\circ}$$ , and $m \angle D= $89^{\circ}$$ 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 $\overline{UD}$ to form two triangles $\bigtriangleup QUD$ and $\bigtriangleup ADU$ . By Statement 1, both triangles are right with congruent legs $\overline{QU}$, $\overline{AD}$ , and congruent hypotenuses, both being the same segment $\overline{UD}$ . By the Hypotenuse Leg Theorem, $\bigtriangleup QUD \cong \bigtriangleup ADU$ . By congruence, $\overline{QD}$ \cong $\overline{AU}$ . The quadrilateral, having two sets of congruent opposite sides, is a parallelogram; a parallelogram with right angles is a rectangle.

Is parallelogram a rectangle?

Statement 1: $\angle A = $90^{\circ }$$

Statement 2:

Tap to see back →

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.

Given a quadrilateral , can a circle be circumscribed about it?

Statement 1: Quadrilateral is not a rectangle.

Statement 2: $m\angle Q = $120^{\circ }$, m \angle A = $50^{\circ }$$

Tap to see back →

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: $m\angle Q = $120^{\circ }$, m\angle U = $120^{\circ }$, m \angle A = $60^{\circ }$, m \angle D = $60^{\circ }$$

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

Case 2: $m\angle Q = $120^{\circ }$, m\angle U = $110^{\circ }$, m \angle A = $70^{\circ }$, m \angle D = $60^{\circ }$$

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.

Given a quadrilateral , can a circle be circumscribed about it?

Statement 1: Quadrilateral is an isosceles trapezoid.

Statement 2: $m\angle Q = $120^{\circ }$, m \angle A = $60^{\circ }$$

Tap to see back →

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 $\angle Q, \angle U$ and $\angle A, \angle D$ are the pairs of base angles.

Then, since base angles are congruent, $m\angle Q =m \angle U$ and $m\angle A =m \angle D$ . Since the bases of a trapezoid are parallel, from the Same-Side Interior Angles Theorem, $\angle Q$ and $\angle D$ are supplementary, and, subsequently, so are $\angle Q$ and $\angle A$ , as well as $\angle U$ and $\angle D$ .

If $m\angle Q = $120^{\circ }$, m \angle A = $60^{\circ }$$ , then $\angle Q$ and $\angle A$ form a supplementary pair, as their measures total ; since the measures of the angles of a quadrilateral total , the measures of $\angle U$ and $\angle D$ 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.

Given Parallelogram .

True or false: $\overline{AC}$ \perp $\overline{BD}$

Statement 1:

Statement 2: $m \angle ABC = $45^{\circ }$$

Tap to see back →

$\overline{AC}$ and $\overline{BD}$ , 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, $\overline{AC}$ \perp $\overline{BD}$ .

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

Are the diagonals of Quadrilateral perpendicular?

(a)

(b)

Tap to see back →

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.

Quadrilateral is inscribed in a circle.

What is $m \angle Q$ ?

Statement 1: $m \angle A = 140 ^{\circ }$

Statement 2: $m \widehat{UAD} = 80 ^{\circ }$

Tap to see back →

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

$m \angle Q = $180^{\circ }$ - m \angle A = $180^{\circ }$ - $140^{\circ }$ = $40^{\circ }$$

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

$m \angle Q = $\frac{1}{2}$\cdot m \widehat{UAD} = $\frac{1}{2}$\cdot 80 ^{\circ } = 40 ^{\circ }$

The above shows Parallelogram . Is it a rectangle?

Statement 1: $m \angle 1 = 90 ^{\circ }$

Statement 2:

Tap to see back →

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 $m \angle 1 = $90^{\circ }$$ , then, since $\angle 1$ and $\angle BCD$ form a linear pair, $\angle BCD$ is right.

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

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

Refer to the above figure. You are given that Polygon is a parallelogram but not that it is a rectangle. Is it a rectangle?

Statement 1: $\angle BAD \cong \angle 1$

Statement 2: $\angle DAC$ and $\angle DCA$ are complementary angles.

Tap to see back →

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

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

Assume Statement 2 alone - that $\angle DAC$ and $\angle DCA$ are complementary angles, or, equivalently, $m \angle DAC + m \angle DCA = 90 ^{\circ }$ . Since the angles of a triangle have measures that add up to , the third angle of $\Delta DAC$ , which is $\angle D$ , measures $90 ^{\circ }$ , and is a right angle.

Either statement alone proves $\angle D$ a right angle and subsequently proves a rectangle.

True or false: Quadrilateral is a rectangle.

Statement 1: $\bigtriangleup QUA \cong \bigtriangleup ADQ$

Statement 2: $\bigtriangleup QUA \cong \bigtriangleup UQD$

Tap to see back →

Assume Statement 1 alone. By congruence, $\overline{QU}$ \cong $\overline{AD}$ and $\overline{QD}$ \cong $\overline{AU}$ , 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 $\overline{QD}$ \cong $\overline{UA}$ , but no clue is provided as to the lengths of opposite sides $\overline{Q U}$ and $\overline{DA}$ . Also, $\angle QUA \cong \angle UQD$ , 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 $\overline{Q U}$ and $\overline{DA}$ .

Assume both statements are true. Quadrilateral is a parallelogram as a consequence of Statement 1. Since $\angle QUA$ and $\angle UQD$ 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.

True or false: Quadrilateral is a rectangle.

Statement 1: $\angle Q$ and $\angle A$ are right angles.

Statement 2: $\overline{QU}$ \cong $\overline{AD}$

Tap to see back →

Statement 1 alone is insufficient to answer the question; A quadrilateral in which $\angle Q$ and $\angle A$ are right angles, $m \angle U = $91^{\circ}$$ , and $m \angle D= $89^{\circ}$$ 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 $\overline{UD}$ to form two triangles $\bigtriangleup QUD$ and $\bigtriangleup ADU$ . By Statement 1, both triangles are right with congruent legs $\overline{QU}$, $\overline{AD}$ , and congruent hypotenuses, both being the same segment $\overline{UD}$ . By the Hypotenuse Leg Theorem, $\bigtriangleup QUD \cong \bigtriangleup ADU$ . By congruence, $\overline{QD}$ \cong $\overline{AU}$ . The quadrilateral, having two sets of congruent opposite sides, is a parallelogram; a parallelogram with right angles is a rectangle.