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ISEE Upper Level Math : Squares

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Example Questions

Example Question #11 : Squares

Side shown in the diagram of square ABCD below is equal to 21cm. What is the area of ABCD ?

Possible Answers:

$210\ cm^{2}$

Cannot be determined

$42\ cm^{2}$

$84\ cm^{2}$

$441\ cm^{2}$

Correct answer:

$441\ cm^{2}$

Explanation:

To find the area of a quadrilateral, multiply length times width. In a square, since all sides are equal, is both the length and width.

$a^{2}=21\times21=441$

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Example Question #2 : How To Find The Area Of A Square

If Amy is carpeting her living room, which meaures feet by feet, how many square feet of carpet will she need?

Possible Answers:

$22 ft^{2}$

$44 ft^{2}$

$120 ft^{2}$

$100 ft^{2}$

$144 ft^{2}$

Correct answer:

$120 ft^{2}$

Explanation:

To find the area of the floor, multiply the length of the room by the width (which is the same forumla used to find the area of a square). The equation can be written: $A=l\cdot w$

Substitute feet for and feet for :

$A=10\cdot 12$

Amy will need $120 ft^{2}$ of carpet.

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Example Question #2 : How To Find The Area Of A Square

A rectangle and a square have the same perimeter. The rectangle has length 100 centimeters and width centimeters. Give the area of the square.

Possible Answers:

$19,600 \textrm{ cm}^{2}$

$4,900 \textrm{ cm}^{2}$

$8,000 \textrm{ cm}^{2}$

$4,000 \textrm{ cm}^{2}$

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

Correct answer:

$4,900 \textrm{ cm}^{2}$

Explanation:

The perimeter of the rectangle is

$P = 2L + 2W = 2\cdot100 +2\cdot40 = 200 + 80 = 280$ centimeters.

This is also the perimeter of the square, so divide this by to get its sidelength:

$s = \frac{P}{4} = 280 \div 4 = 70$ centimeters.

The area is the square of this, or $A = s^{2} = 70^{2} = 4,900$ square centimeters.

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Example Question #11 : Squares

Four squares have sidelengths 4 inches, 8 inches, 12 inches, and 16 inches. What is the average of their areas?

Possible Answers:

$200\textrm{ in} ^{2}$

$64\textrm{ in} ^{2}$

$144\textrm{ in} ^{2}$

$100\textrm{ in} ^{2}$

$120\textrm{ in} ^{2}$

Correct answer:

$120\textrm{ in} ^{2}$

Explanation:

The areas of the four squares can be calculated by squaring their sidelengths. Add these areas, then divide by 4:

$\frac{4^{2}+8^{2}+12^{2}+16^{2}}{4} = \frac{16+64+144+256}{4} = \frac{480}{4} = 120$ square inches

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Example Question #12 : Squares

Which of the following is equal to the area of a square with sidelength $1 \frac{1}{6}$ yards?

Possible Answers:

$1,764 \textrm{ in}^{2}$

$2,304 \textrm{ in}^{2}$

$2,025 \textrm{ in}^{2}$

$1,225 \textrm{ in}^{2}$

Correct answer:

$1,764 \textrm{ in}^{2}$

Explanation:

Multiply the sidelength by 36 to convert from yards to inches:

$1 \frac{1}{6} \times36 = \frac{7}{6} \times \frac{36}{1} = \frac{7}{1} \times \frac{6}{1} = 42\ inches$

Square this to get the area:

$42^{2} = 1,764$ square inches

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Example Question #21 : Quadrilaterals

What is the area of a square in which the length of one side is equal to $\sqrt{\frac{1}{4}+\frac{1}{3}}$ ?

Possible Answers:

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{5}{12}$

$\frac{7}{12}$

Correct answer:

$\frac{7}{12}$

Explanation:

The area of a square is equal to the product of one side multiplied by another side. Therefore, the area will be equal to:

$\sqrt{\frac{1}{4}+\frac{1}{3}}\cdot \sqrt{\frac{1}{4}+\frac{1}{3}}=\frac{1}{4}+\frac{1}{3}$

The next step is to convert the fractions being added together to a form in which they have a common denominator. This gives us:

$\frac{3}{12}+\frac{4}{12}=\frac{7}{12}$

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Example Question #15 : Squares

One of the sides of a square on the coordinate plane has its endpoints at the points with coordinates (6, 2) and (-5, 1) . What is the area of this square?

Possible Answers:

120

122

Correct answer:

122

Explanation:

The length of a segment with endpoints (6, 2) and (-5, 1) can be found using the distance formula with $x_{1} = 6$ , $y_{1} = 2$ , $x_{2} = -5$ , $y_{2} = 1$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$= \sqrt{(6 - (-5))^{2}+(2-1)^{2}}$

$= \sqrt{11^{2}+1^{2}}$

$= \sqrt{121+1}$

$= \sqrt{122}$

This is the length of one side of the square, so the area is the square of this, or 122.

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Example Question #5 : How To Find The Area Of A Square

One of the sides of a square on the coordinate plane has its endpoint at the points with coordinates (A, B) and (-B, A) , where and are both positive. Give the area of the square in terms of and .

Possible Answers:

4AB

$2A^{2} +2 B^{2}$

$2A^{2} -4AB +2 B^{2}$

$2A^{2} +4AB +2 B^{2}$

Correct answer:

$2A^{2} +2 B^{2}$

Explanation:

The length of a segment with endpoints (A, B) and (-B, A) can be found using the distance formula as follows:

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{((-B)-A)^{2}+(A-B)^{2}}$

$d = \sqrt{(-B-A)^{2}+(A-B)^{2}}$

$d = \sqrt{(-1)^{2}(B+A)^{2}+(A-B)^{2}}$

$d = \sqrt{1 \cdot (B^{2} +2AB +A^{2})+(A^{2}-2AB+B^{2})}$

$d = \sqrt{ (B^{2} +2AB +A^{2})+(A^{2}-2AB+B^{2})}$

$d = \sqrt{ 2A^{2} +2 B^{2}}$

This is the length of one side of the square, so the area is the square of this, or $2A^{2} +2 B^{2}$ .

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Example Question #6 : How To Find The Area Of A Square

One of the vertices of a square is at the origin. The square has area 13. Which of the following could be the vertex of the square opposite that at the origin?

Possible Answers:

(6,0)

(3, 3)

(4, 2)

(5, 1)

Correct answer:

(5, 1)

Explanation:

Since a square is a rhombus, one way to calculate the area of a square is to take half the square of the length of a diagonal. If we let be the length of each diagonal, then

$\frac{1}{2} d^{2} = A$

$\frac{1}{2} d^{2} = 13$

$d^{2} =26$

$d = \sqrt{26}$

Therefore, we want to choose the point that is $\sqrt{26}$ units from the origin. Using the distance formula, we see that (5, 1) is such a point:

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$= \sqrt{(5-0)^{2}+(1-0)^{2}}$

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

$= \sqrt{25+1}$

$= \sqrt{26}$

Of the other points:

(6,0) :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$= \sqrt{(6-0)^{2}+(0-0)^{2}}$

$= \sqrt{6^{2}+0^{2}}$

$= \sqrt{36+0}$

$= \sqrt{36 }$

(4, 2) :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$= \sqrt{(4-0)^{2}+(2-0)^{2}}$

$= \sqrt{4^{2}+2^{2}}$

$= \sqrt{16+4}$

$= \sqrt{20}$

(3, 3) :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$= \sqrt{(3-0)^{2}+(3-0)^{2}}$

$= \sqrt{3^{2}+3^{2}}$

$= \sqrt{9+9}$

$= \sqrt{18}$

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Example Question #22 : Quadrilaterals

One of your holiday gifts is wrapped in a cube-shaped box.

If one of the edges has a length of 6 inches, what is the area of one side of the box?

Possible Answers:

18in^2

12in^2

36 in

36 in^2

Correct answer:

36 in^2

Explanation:

One of your holiday gifts is wrapped in a cube-shaped box.

If one of the edges has a length of 6 inches, what is the area of one side of the box?

We are asked to find the area of one side of a cube, in other words, the area of a square.

We can find the area of a square by squaring the length of the side.

$A_{square}=s^2=(6in)^2=36in^2$

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Squares

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #11 : Squares

Example Question #2 : How To Find The Area Of A Square

Example Question #2 : How To Find The Area Of A Square

Example Question #11 : Squares

Example Question #12 : Squares

Example Question #21 : Quadrilaterals

Example Question #15 : Squares

Example Question #5 : How To Find The Area Of A Square

Example Question #6 : How To Find The Area Of A Square

Example Question #22 : Quadrilaterals

All ISEE Upper Level Math Resources

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