GMAT Quantitative - Calculating the area of a square

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

$N^{2}-12$

$N^{2}-9$

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: $4 \left (N - 3 \right ) = 4N - 12$

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?

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

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

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

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

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

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:

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

Squares

Note: Figure NOT drawn to scale

Refer to the above figure, which shows Square and Square . and Square has area 25. Give the area of Square .

$25+4\sqrt{6}$

$25 + 2\sqrt{6}$

$25 + 8\sqrt{6}$

Explanation

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

Specifically, , and, as given, .

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 is the area of Square plus the areas of the four triangles, or $25 + 4\sqrt{6}$ .

What polynomial represents the area of Square if ?

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

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

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

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

$4t ^{2} +25$

Explanation

As a square, 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 :

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

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

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

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

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

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?

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

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

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

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

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

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.

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

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

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

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

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

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

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:

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

square inches

Squares

Note: Figure NOT drawn to scale

Refer to the above figure, which shows Square and Square . and Square has area 49. Give the area of Square .

$49-2\sqrt{6}$

$49-4\sqrt{6}$

$49- \sqrt{6}$

Explanation

Square has area 49, so each of its sides has as its length the square root of 49, or 7. Each side of Square is therefore a hypotenuse of a right triangle with legs 1 and , 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 .

The perimeter of a square is the same as the circumference of a circle with area 100. What is the area of the square?

$25 \pi$

$\frac{100 }{\pi}$

$\frac{25 \pi^{2}}{4}$

$50 \sqrt{\pi}$

$\frac{200 \sqrt{\pi}}{\pi}$

Explanation

The formula for the area of a circle is

$A = \pi r^{2}$ .

If the area is 100, the radius is as follows:

$\pi r^{2} = 100$

$r^{2} = \frac{100}{\pi}$

$r = \sqrt{\frac{100}{\pi}}= \frac{10}{\sqrt{\pi}}$

The circle has circumference $2 \pi$ times its radius, or

$C= \frac{10}{\sqrt{\pi}} \cdot 2 \pi = 20 \sqrt{\pi}$

This is also the perimeter of the square, so the sidelength of the square is one-fourth this, or

$S = \frac{1}{4} \cdot 20 \sqrt{\pi} = 5 \sqrt{\pi}$

The area of the square is the square of this, or

$A'= \left (5 \sqrt{\pi} \right )^{2}= 25 \pi$

Squares

Note: Figure NOT drawn to scale

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

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

Explanation

To make this easier, assume that and ; the results generalize.

Each side of Square has length 8, so the area of Square 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 has area four times this subtracted from the area of Square , or