GMAT Quantitative - Squares

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A circle is inscribed in a square. The area of the circle is $2\pi$ . What is the area of the square?

$2\sqrt{2}$

$3\sqrt{2}$

Explanation

Since we know the area of the circle, we can tell that: $\pi \cdot r^{2}= 2\pi$ . Where is the radius of the circle.

The length of a side of the square will be since the diameter of the circle is the same length as the side length of the square.

Finally we can calculate the area of the square which will be $(2r)^{2}$ . $r=\sqrt{2}$ so the area will be , which is our final answer.

Two squares in the same plane have the same center. The length of one side of the smaller square is 10; the area of the region between the squares is 60. Give the length of one side of the larger square.

$4\sqrt{10}$

$8\sqrt{2}$

Explanation

Let be the length of one side of the larger square. Then the larger square has area $S^{2}$ ; the smaller square has area $10^{2} = 100$ . The area of the region between them, 60, is their difference:

$S^{2} - 100 = 60$

$S^{2} = 160$

$S = \sqrt{160} = \sqrt{16} \cdot \sqrt{10} = 4\sqrt{10}$

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A circle is inscribed in a square. The square has an area of , what is the perimeter of the circle?

$\sqrt{3}\pi$

$\sqrt{3}$

$3\pi$

$\pi$

$2\sqrt{3}\pi$

Explanation

The square as an area of 3. From that we can figure out the length of a side of the square, which is the same size as the diameter of the circle.

$A=s^2 \rightarrow 3=s^2$

$\sqrt3=s$

From the diameter of the circle we can find out, the perimeter of the circle given by the formula: $d\pi$ , where is the length of the diamter.

So $d=\sqrt{3}$ and therefore, the final answer is $\sqrt{3}\pi$ .

Two squares in the same plane have the same center. The length of one side of the larger square is 10; the area of the region between the squares is 60. Give the length of one side of the smaller square.

$2 \sqrt{10}$

$4\sqrt{5}$

$6 \sqrt{2}$

Explanation

Let be the length of one side of the smaller square. Then the smaller square has area $S^{2}$ ; the larger square has area $10^{2} = 100$ . The area of the region between them, 60, is their difference:

$60 = 100 - S^{2}$

$60+ S^{2} - 60 = 100 - S^{2}+ S^{2} - 60$

$S^{2} = 40$

$S = \sqrt{40} = \sqrt{4} \cdot \sqrt{10} = 2 \sqrt{10}$

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

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A circle is inscribed in a square. The diagonal of the square is $2\sqrt{2}$ , what is the area of the circle?

$\pi$

$\sqrt{2}\pi$

$2\pi$

$2\sqrt{2}\pi$

Explanation

From what we are told, we can figure out the side of the square with this simple formula: $d=s\sqrt{2}$ , where is the diagonal of the square and is the length of the side. So the length of the side of our square will be 2. Therefore the radius of the circle will be half that, or 1. Now that we now the radius, we can calculate the area of the circle $\pi r^{2}$ , where is the radius of the circle.

$A=\pi(1)^2=\pi$

Find the length of the diagonal of a square whose side length is .

$4\sqrt2$

$16\sqrt2$

Explanation

To find the length of the diagonal, you can either use the pythagorean theorem or realize that the diagonal of a square is really the hypotenuse of a right isosceles triangle. Therefore, using the short cut, you know that the diagonal is the side length times $\sqrt2$ . Thus, the answer is: $4\sqrt2$

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, 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:

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

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.

Squares

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