GMAT Math : Calculating an angle in a quadrilateral

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Example Question #1 : Calculating An Angle In A Quadrilateral

Parallelogram

Note: Diagram is NOT drawn to scale.

Refer to the above diagram.

Any of the following facts alone would be enough to prove that $\textrm{Quad} \; ABCD$ is not a parallelogram, EXCEPT:

Possible Answers:

Any one of these facts alone would prove that $\textrm{Quad} \; ABCD$ is not a parallelogram.

$m \angle A = 114^{\circ}$

BC > 12

$m \angle D <76^{\circ}$

CD = 12

Correct answer:

BC > 12

Explanation:

Opposite sides of a parallelogram are congruent; if CD = 12 , then $CD \neq AB$ , violating this condition.

Consecutive angles of a parallelogram are supplementary; if $m \angle A = 114^{\circ}$ , then $m \angle A + m \angle B= 114 + 76 = 190$ , violating this condition.

Opposite angles of a parallelogram are congruent; if $m \angle D <76^{\circ}$ , then $m \angle D \neq m \angle B$ , violating this condition.

Adjacent sides of a parallelogram, however, may or may not be congruent, so the condition that BC > 12 would not by itself prove that the quadrilateral is not a parallelogram.

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Example Question #1 : Calculating An Angle In A Quadrilateral

Which of the following can not be the measures of the four interior angles of a quadrilateral?

Possible Answers:

All four of the other choices fit the conditions.

$80^{\circ },80^{\circ },100^{\circ },100^{\circ }$

$60^{\circ },100^{\circ },100^{\circ },100^{\circ }$

$75^{\circ },85^{\circ },95^{\circ },105^{\circ }$

$75^{\circ },75^{\circ },75^{\circ },125^{\circ }$

Correct answer:

$75^{\circ },75^{\circ },75^{\circ },125^{\circ }$

Explanation:

The four interior angles of a quadrilateral measure a total of $360^{\circ }$ , so we test each group of numbers to see if they have this sum.

75+85+95+105 = 360

80+80+100+100 = 360

60+100+100+100 = 360

75 + 75 + 75 + 125 = 350

This last group does not have the correct sum, so it is the correct choice.

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Example Question #1 : Calculating An Angle In A Quadrilateral

A circle can be circumscribed about each of the following figures except:

Possible Answers:

A rectangle twice as long as it is wide

A rhombus with a $60^{\circ }$ angle

An isosceles trapezoid with one base three times as long as the other

A right scalene triangle

An isosceles triangle with its base one-half as long as either leg

Correct answer:

A rhombus with a $60^{\circ }$ angle

Explanation:

A circle can be circumscribed about any triangle regardless of its sidelengths or angle measures, so we can eliminate the two triangle choices.

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; this can be proved by the fact that base angles are congruent, and by the Same-Side Interior Angles Statement. For a parallelogram to have this characteristic, since opposite angles are congruent also, all angles must measure $90^{\circ }$ ; the rectangle fits this description, but the rhombus does not.

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Example Question #3 : Calculating An Angle In A Quadrilateral

Rhombus RHOM has two diagonals that intersect at point ; $\angle RHO = 80 ^{\circ }$ .

What is $m \angle RPH$ ?

Possible Answers:

$40 ^{\circ }$

$90 ^{\circ }$

$80 ^{\circ }$

$100 ^{\circ }$

$50 ^{\circ }$

Correct answer:

$90 ^{\circ }$

Explanation:

The diagonals of a rhombus always intersect at right angles, so $m \angle HPO = 90 ^{\circ }$ . The measures of the interior angles of the rhombus are irrelevant.

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Example Question #241 : Geometry

Quadrilateral QUAD is inscribed in circle $\bigodot$ . $m \angle A = 100 ^{\circ }$ . What is $m \angle Q$ ?

Possible Answers:

$40^{\circ }$

$160 ^{\circ }$

$100 ^{\circ }$

$50 ^{\circ }$

$80 ^{\circ }$

Correct answer:

$80 ^{\circ }$

Explanation:

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

$m \angle Q = 180^{\circ } - m \angle A = 180^{\circ } - 100^{\circ } = 80^{\circ }$

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Example Question #41 : Other Quadrilaterals

Rhombus

Note: Figure NOT drawn to scale.

The above figure is of a rhombus and one of its diagonals. What is equal to?

Possible Answers:

x = 114

x = 138

x = 156

x = 132

Not enough information is given to answer the question.

Correct answer:

x = 132

Explanation:

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:

$\left (180 - 2 \times 24 \right )^{\circ } = \left (180 -48 \right )^{\circ } = 132^{\circ }$

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Example Question #1 : Calculating An Angle In A Quadrilateral

Rectangle

Refer to the above figure. You are given that Polygon ABCD is a parallelogram, but NOT that it is a rectangle.

Which of the following statements is not enough to prove that Polygon ABCD is also a rectangle?

Possible Answers:

$\overline{BC} \perp \overline{DC}$

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

$\angle BAC \cong \angle ACD$

$\angle BAC$ and $\angle CAD$ are complementary angles

$(AD)^{2} + (DC)^{2} = (AC)^{2}$

Correct answer:

$\angle BAC \cong \angle ACD$

Explanation:

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

If $\overline{BC} \perp \overline{DC}$ , then by definition of perpendicular lines, $\angle BCD$ is right.

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

If $(AD)^{2} + (DC)^{2} = (AC)^{2}$ , then, by the Converse of the Pythagorean Theorem, $\Delta ADC$ is a right triangle with right angle $\angle D$ .

If $\angle BAC$ and $\angle CAD$ are complementary angles, then, since $m \angle BAC + m \angle CAD =m \angle BA D$

$m \angle BAD = m \angle BAC + m \angle CAD = 90 ^{\circ }$ , making $m \angle BAD$ right.

However, since, by definition of a parallelogram, $\overline{AB } \parallel \overline{ CD }$ , by the Alternate Interior Angles Theorem, $\angle BAC \cong \angle ACD$ regardless of whether the parallelogram is a rectangle or not.

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Example Question #2 : Calculating An Angle In A Quadrilateral

Two angles of a parallelogram measure $\left ( x + 70 \right )^{\circ }$ and $\left ( 2x\right +5) ^{\circ }$ . What are the possible values of ?

Possible Answers:

$x = 35 \textrm{ or } x = 95$

x = 35

$x = 35 \textrm{ or } x = 65$

x = 65

x = 95

Correct answer:

$x = 35 \textrm{ or } x = 65$

Explanation:

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

2x +5 = x + 70

2x +5 - x - 5 = x + 70 - x - 5

x = 65

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

$\left ( 2x +5 \right )+\left ( x + 70 \right ) = 180$

3x+ 75 = 180

3x+ 75 -75 = 180 -75

3x = 105

$3x \div 3 = 105\div 3$

x = 35

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GMAT Math : Calculating an angle in a quadrilateral

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Calculating An Angle In A Quadrilateral

Example Question #1 : Calculating An Angle In A Quadrilateral

Example Question #1 : Calculating An Angle In A Quadrilateral

Example Question #3 : Calculating An Angle In A Quadrilateral

Example Question #241 : Geometry

Example Question #41 : Other Quadrilaterals

Example Question #1 : Calculating An Angle In A Quadrilateral

Example Question #2 : Calculating An Angle In A Quadrilateral

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