GMAT Math : Quadrilaterals

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Example Questions

Example Question #241 : Geometry

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 = 138

Not enough information is given to answer the question.

x = 156

x = 114

x = 132

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 #51 : Other Quadrilaterals

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

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

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

x = 95

x = 65

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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Example Question #1 : Calculating The Perimeter Of A Quadrilateral

Quad

Note: Figure NOT drawn to scale

What is the perimeter of Quadrilateral QUAD , above?

Possible Answers:

Correct answer:

Explanation:

By the Pythagorean Theorem,

$\left (AD \right )^{2} +\left ( AU\right ) ^{2} =\left ( DU\right ) ^{2}$

$\left (AD \right )^{2} +4 ^{2} =5 ^{2}$

$\left (AD \right )^{2} +16 =25$

$\left (AD \right )^{2} +16 -16 =25 -16$

$\left (AD \right )^{2} =9$

AD = 3

Also by the Pythagorean Theorem,

$\left ( QU\right ) ^{2} + \left ( DU\right ) ^{2}=\left ( DQ\right ) ^{2}$

$\left ( QU\right ) ^{2} + 5^{2}=13^{2}$

$\left ( QU\right ) ^{2} + 25 =169$

$\left ( QU\right ) ^{2} + 25 -25=169 -25$

$\left ( QU\right ) ^{2} =144$

QU = 12

The perimeter of Quadrilateral QUAD is

QU + UA + AD + DQ = 12 + 4 + 3 + 13 = 32

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Example Question #131 : Quadrilaterals

What is the perimeter of Rhombus ABCD ?

Statement 1: AB = 10

Statement 2: Rhombus ABCD has area .

Possible Answers:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Correct answer:

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Explanation:

A rhombus has four congruent sides. Statement 1 gives the length of one of them, so that length can be multiplied by to yield the perimeter.

The area of a rhombus alone has no bearing on its perimeter, so Statement 2 alone is insufficient.

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Example Question #245 : Geometry

Isosceles trapezoid FGHJ has a side length of $5 \:in$ . The length of one of its bases (Base A) is equal to three times the amount of seven inches fewer than the length of the other base (Base B). If Base B has a length of $17 \:in$ , what is the perimeter of FGHJ ?

Possible Answers:

$61 \:in$

$30 \:in$

$57 \:in$

$54 \:in$

$27 \:in$

Correct answer:

$57 \:in$

Explanation:

In an isosceles trapezoid, the two non-parallel sides are equal. In this case, they are both $5\:in$ long. To find the perimeter, we also need to know the length of both bases. We are told that Base B is $17 \:in$ . Eliminate any options less than or equal to $27 \:in$ .

To find the length of Base A, we need to translate the following: "The length of one of its bases (Base A) is equal to three times the amount of seven inches fewer than the length of the other base (Base B)." For translating, break the statement down:

Starting with $Base \:B=17 \:in$ ,

"seven inches fewer than the length of the other base (Base B)": $17\:in-7\:in$

"three times the amount of seven inches fewer than the length of the other base (Base B)": $3(17\:in-7\:in)$

$3(17\:in-7\:in)=3(10\:in)=30\:in$

So, Base A is $30 \:in$ long, so the perimeter of isosceles trapezoid FGHJ is:

$30\:in+5\:in+5\:in+17\:in=57 \:in$

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Example Question #2 : Calculating The Perimeter Of A Quadrilateral

Frank is planning on fencing a rectangular field near his house. The longer side of the house is two times four more than the length of the other side. If the shorter side is meters, what is the total length of fence that Frank needs?

Possible Answers:

176 m

116 m

6720 m

120 m

352 m

Correct answer:

352 m

Explanation:

This is a perimeter problem, but first we need to find our side lengths.

The short sides are meters. The long sides are two times 4 more than the short sides. So

"four more" 56+4=60

"two times" $60\cdot2=120$ meters

So our two side lengths are 120 and 56 meters. Find the perimeter by the following:

2(56)+2(120)=112+240=352

So, 352 meters

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Example Question #1 : Calculating The Perimeter Of A Quadrilateral

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Rhombus ACBD has diagonals AB=4 and CD=2 . What is the perimeter of the rhombus?

Possible Answers:

$\sqrt{6}$

$2\sqrt{5}$

$4\sqrt{5}$

$\sqrt{5}$

Correct answer:

$4\sqrt{5}$

Explanation:

The rhombus is a special kind of a parallelogram. Its sides are all of the same length. Therefore, we just need to find one length of this quadrilateral. To do so, we can apply the Pythagorean Theorem on triangle AEC for example, since we know the length of the diagonals. Also, the diagonals intersect at their center. Therefore, triangle AEC has length, AE=2 and CE=1 . Therefore, $AC=\sqrt{1^{2}+2^{2}}$ or $AC= \sqrt{5}$ . The perimeter is then $4\sqrt{5}$ .

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GMAT Math : Quadrilaterals

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #241 : Geometry

Example Question #1 : Calculating An Angle In A Quadrilateral

Example Question #51 : Other Quadrilaterals

Example Question #1 : Calculating The Perimeter Of A Quadrilateral

Example Question #131 : Quadrilaterals

Example Question #245 : Geometry

Example Question #2 : Calculating The Perimeter Of A Quadrilateral

Example Question #1 : Calculating The Perimeter Of A Quadrilateral

All GMAT Math Resources

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