GMAT Math : Calculating the area of a square

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Example Questions

Example Question #1 : Squares

Write, in terms of , the perimeter of a square whose area is $N^{2 } - 6N + 9$

Possible Answers:

4N-12

4N-36

4N-24

$N^{2}-12$

$N^{2}-9$

Correct answer:

4N-12

Explanation:

To find the perimeter of a square given its area, take the square root of the area to find its sidelength; then, multiply that sidelength by 4.

$N^{2 } - 6N + 9$ is a perfect square trinomial, since $\left (\frac{-6}{2} \right )^{2} =(-3)^{2}= 9$

so its square root is $\sqrt{N^{2 } - 6N + 9} = N - 3$ , the sidelength.

Multiply this by 4 to get the perimeter: $4 \left (N - 3 \right ) = 4N - 12$

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Example Question #4 : Squares

If the diagonal of a square room is $\frac{x}{2}$ . What is the area of the room?

Possible Answers:

$\frac{x^2}{8}$

$\frac{x^2}{4}$

$\frac{x^2}{2}$

x^2

Correct answer:

$\frac{x^2}{8}$

Explanation:

Cutting the triangle in half yields a right triangle with the diagonal becoming the hypotenuse and the other two legs being the sides of the square. Using the Pythagorean Theorem, we can solve for the other legs of the triangle.

a^2+b^2=c^2

Since both sides of the square are equal to eachother, a=b , therefore:

a^2+a^2=c^2

$2a^2=\left(\frac{x}{2}\right)^2$

$2a^2= \frac{x^2}{4}$

$a^2=\frac{x^2}{8}$

$a= \frac{x}{\sqrt{8}}$

To find the area of the square:

A=s^2 with leg being one of the sides

$A= \left(\frac{x}{\sqrt{8}}\right)^2$

$A=\frac{x^2}{8}$

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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 #2 : Calculating The Area Of A Square

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:

$225 \:m^2$

$60 \:m^2$

$30 \:m^2$

$120 \:m^2$

$45 \:m^2$

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 #9 : 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:

$49-2\sqrt{6}$

$49-4\sqrt{6}$

$49- \sqrt{6}$

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 #1 : 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 WXYZ has area 25. Give the area of Square ABCD .

Possible Answers:

$25 + 2\sqrt{6}$

$25 + 8\sqrt{6}$

$25+4\sqrt{6}$

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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GMAT Math : Calculating the area of a square

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Squares

Example Question #4 : Squares

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 #2 : Calculating The Area Of A Square

Example Question #9 : Calculating The Area Of A Square

Example Question #1 : Calculating The Area Of A Square

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