Calculating an angle in a quadrilateral - GMAT Quantitative Reasoning

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Question

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:

Answer

Opposite sides of a parallelogram are congruent; if , then , 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 $m \angle D eq m \angle B$ , violating this condition.

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

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Question

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

Answer

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.

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

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Question

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

Answer

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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Question

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

Answer

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

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

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

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

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Question

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

What is $m \angle RPH$ ?

Answer

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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Question

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

Answer

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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Question

Note: Figure NOT drawn to scale.

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

Answer

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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Question

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

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

Answer

To prove that Polygon 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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