GMAT Math : Quadrilaterals

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Example Questions

Example Question #2 : Calculating The Area Of A Square

A square plot of land has perimeter 1,200 feet. Give its area in square yards.

Possible Answers:

$160,000\textup{ yd} ^{2}$

$20,000 \textup{ yd} ^{2}$

$100,000\textup{ yd} ^{2}$

$90,000\textup{ yd} ^{2}$

$10,000 \textrm{ yd} ^{2}$

Correct answer:

$10,000 \textrm{ yd} ^{2}$

Explanation:

The length of one side of the square is $1,200 \div 4 = 300$ feet, or $300 \div 3 = 100$ yards. Square this to get the area in square yards:

$A =100^{2} =10,000$ square yards.

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Example Question #3 : Calculating The Area Of A Square

A square, a regular pentagon, and a regular hexagon have the same sidelength. The sum of their perimeters is one mile. What is the area of the square?

Possible Answers:

$193,600\textrm{ ft}^{2}$

$278,784\textrm{ ft}^{2}$

$123,904\textrm{ ft}^{2}$

$69,696\textrm{ ft}^{2}$

$108,900\textrm{ ft}^{2}$

Correct answer:

$123,904\textrm{ ft}^{2}$

Explanation:

The square, the pentagon, and the hexagon have a total of 15 sides, all of which are of equal length; the sum of the lengths is one mile, or 5,280 feet, so the length of one side of any of these polygons is

$5,280 \div 15 = 352$ feet.

The square has area equal to the square of this sidelength:

$352^{2} = 123,904 \textup{ ft} ^{2}$

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Example Question #4 : Calculating The Area Of A Square

A square and a regular pentagon have the same perimeter. The length of one side of the pentagon is 60 centimeters. What is the area of the square?

Possible Answers:

$3,600 \textup{ cm}^{ 2}$

$5,625 \textup{ cm}^{ 2}$

$5,175 \textup{ cm}^{ 2}$

$7,500 \textup{ cm}^{ 2}$

$6,400 \textup{ cm}^{ 2}$

Correct answer:

$5,625 \textup{ cm}^{ 2}$

Explanation:

The regular perimeter has sidelength 60 centimeters and therefore perimeter $5 \times 60 = 300$ centimeters. The square has as its sidelength $300 \div 4 = 75$ centimeters and area $75 ^{2} = 5,625$ square centimeters.

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Example Question #2 : Calculating The Area Of A Square

Six squares have sidelengths 8 inches, 1 foot, 15 inches, 20 inches, 2 feet, and 25 inches. What is the sum of their areas?

Possible Answers:

$416\textrm{ in}^{2}$

$1,319\textrm{ in}^{2}$

$2,034 \textrm{ in}^{2}$

$4,068\textrm{ in}^{2}$

$10,816\textrm{ in}^{2}$

Correct answer:

$2,034 \textrm{ in}^{2}$

Explanation:

The areas of the squares are the squares of the sidelengths, so add the squares of the sidelengths. Since 1 foot is equal to 12 inches and 2 feet are equal to 24 inches, the sum of the areas is:

$8^{2} + 12 ^{2} + 15 ^{2} + 20 ^{2} + 24^{2} + 25^{2}$

= 64 + 144 + 225 + 400 + 576 + 625

=2,034 square inches

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Example Question #1 : Calculating The Area Of A Square

What polynomial represents the area of Square ABCD if AC = 2t + 5 ?

Possible Answers:

$4t ^{2} +25$

$2t ^{2} +\frac{ 25 }{2}$

$4t ^{2} +20t +25$

$2t ^{2} +10t +\frac{ 25 }{2}$

$2t ^{2} +10t + 25$

Correct answer:

$2t ^{2} +10t +\frac{ 25 }{2}$

Explanation:

As a square, ABCD is also a rhombus. The area of a rhombus is half the product of the lengths of its diagonals, one of which is $\overline{AC }$ . Since the diagonals are congruent, this is equal to half the square of :

$\frac{1}{2} \left (AC \right) ^{2}$

$= \frac{1}{2} \left (2t + 5 \right) ^{2}$

$= \frac{1}{2}\left [ \left (2t \right) ^{2} + 2\left (2t \right) \cdot 5 + 5^{2} \right ]$

$= \frac{1}{2}\left ( 4t ^{2} +20t + 25 \right )$

$= 2t ^{2} +10t +\frac{ 25 }{2}$

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Example Question #401 : Gmat Quantitative Reasoning

Given square FGHI, answer the following

Square1

If square FGHI represents the surface of an ancient arena discovered by archaeologists, what is the area of the arena?

Possible Answers:

$120 \:m^2$

$30 \:m^2$

$60 \:m^2$

$45 \:m^2$

$225 \:m^2$

Correct answer:

$225 \:m^2$

Explanation:

This problem requires us to find the area of a square. Don't let the story behind it distract you, it is simply an area problem. Use the following equation to find our answer:

$Area=s^2=(15\:m)^2=225\:m^2$

is the length of one side of the square; in this case we are told that it is $15\:m$ , so we can solve accordingly!

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Example Question #2 : Calculating The Area Of A Square

Squares

Note: Figure NOT drawn to scale

Refer to the above figure, which shows Square ABCD and Square WXYZ . BW = 1 and Square ABCD has area 49. Give the area of Square WXYZ .

Possible Answers:

$49-4\sqrt{6}$

$49-2\sqrt{6}$

$49- \sqrt{6}$

Correct answer:

Explanation:

Square ABCD has area 49, so each of its sides has as its length the square root of 49, or 7. Each side of Square WXYZ is therefore a hypotenuse of a right triangle with legs 1 and 7-1 = 6 , so each sidelength, including , can be found using the Pythagorean Theorem:

$WX = \sqrt{(WB)^{2}+(BX)^{2}}$

$WX = \sqrt{1^{2}+6^{2}} = \sqrt{1 +36} = \sqrt{37}$

The square of this, which is 37, is the area of Square WXYZ .

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Example Question #51 : Quadrilaterals

Squares

Note: Figure NOT drawn to scale

Refer to the above figure, which shows Square ABCD and Square WXYZ . BW = 1 and Square WXYZ has area 25. Give the area of Square ABCD .

Possible Answers:

$25+4\sqrt{6}$

$25 + 8\sqrt{6}$

$25 + 2\sqrt{6}$

Correct answer:

$25+4\sqrt{6}$

Explanation:

Square WXYZ has area 25, so each side has length the square root of 25, or 5.

Specifically, WX = 5 , and, as given, BW = 1 .

Since $\bigtriangleup WBX$ is a right triangle with hypotenuse $\overline{WX}$ and legs $\overline{BW}$ and $\overline{BX}$ , can be found using the Pythagorean Theorem:

$BX = \sqrt{(WX)^{2}-(BW)^{2}}$

$= \sqrt{5^{2}-1^{2}}$

$= \sqrt{25-1}$

$= \sqrt{24}$

$=\sqrt{4} \cdot \sqrt{6}$

$= 2\sqrt{6}$

The area of $\bigtriangleup WBX$ is

$\frac{1}{2} \cdot BW \cdot BX = \frac{1}{2} \cdot 1 \cdot 2 \sqrt{6} = \sqrt{6}$

Since all four triangles, by symmetry, are congruent, all have this area. the area of Square ABCD is the area of Square WXYZ plus the areas of the four triangles, or $25 + 4\sqrt{6}$ .

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Example Question #401 : Problem Solving Questions

Squares

Note: Figure NOT drawn to scale

Refer to the above figure, which shows Square ABCD and Square WXYZ . The ratio of to is 13 to 2. What is the ratio of the area of Square ABCD to that of Square WXYZ ?

Possible Answers:

17:15

225:221

225:173

225:169

15:13

Correct answer:

225:173

Explanation:

To make this easier, assume that BX = 13 and BW=CX = 2 - the reasoning generalizes. Then Square ABCD has sidelength 15 and area $15^{2}= 225$ . The sidelength of Square WXYZ , each side being a hypotenuse of a right triangle with legs 2 and 13, is

$\sqrt{2^{2}+13^{2}}= \sqrt{4+169} = \sqrt{173}$ .

The square of this, 173, is the area of Square WXYZ .

The ratio is 225:173 .

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Example Question #171 : Geometry

Squares

Note: Figure NOT drawn to scale

Refer to the above figure, which shows Square ABCD and Square WXYZ . The ratio of to is 7 to 1.

Which of these responses comes closest to what percent the area of Square WXYZ is of that of Square ABCD ?

Possible Answers:

$75 \%$

$90 \%$

$85 \%$

$70 \%$

$80 \%$

Correct answer:

$80 \%$

Explanation:

To make this easier, assume that BX = 7 and BW = XC = 1 ; the results generalize.

Each side of Square ABCD has length 8, so the area of Square ABCD is 64.

Each of the four right triangles has legs 7 and 1, so each has area $\frac{1}{2} \cdot 1 \cdot 7 = 3\frac{1}{2}$ ; Square WXYZ has area four times this subtracted from the area of Square ABCD , or

$64 - 4 \cdot 3 \frac{1}{2} = 50$ .

The area of Square WXYZ is

$\frac{50}{64} \times 100 \% = 78 \frac{1}{8} \%$

of that of Square ABCD .

Of the five choices, 80% comes closest.

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GMAT Math : Quadrilaterals

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #2 : Calculating The Area Of A Square

Example Question #3 : Calculating The Area Of A Square

Example Question #4 : Calculating The Area Of A Square

Example Question #2 : Calculating The Area Of A Square

Example Question #1 : Calculating The Area Of A Square

Example Question #401 : Gmat Quantitative Reasoning

Example Question #2 : Calculating The Area Of A Square

Example Question #51 : Quadrilaterals

Example Question #401 : Problem Solving Questions

Example Question #171 : Geometry

All GMAT Math Resources

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