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ISEE Upper Level Quantitative : Squares

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Example Questions

Example Question #221 : Plane Geometry

The lengths of the sides of ten squares form an arithmetic sequence. One side of the smallest square measures sixty centimeters; one side of the second-smallest square measures one meter.

Give the area of the largest square, rounded to the nearest square meter.

Possible Answers:

22 square meters

16 square meters

18 square meters

20 square meters

24 square meters

Correct answer:

18 square meters

Explanation:

Let $a_{n}$ be the lengths of the sides of the squares in meters. $a_{1} = 60 \div 100 = 0.6$ and $a_{2} = 1$ , so their common difference is

d =1 - 0.6 = 0.4

The arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

The length of a side of the largest square - square 10 - can be found by substituting $a_{1} = 0.6, n= 10, d = 0.4$ :

$a_{10} =0.6+ (10-1) 0.4 = 0.6 + 9 (0.4) = 0.6 + 3.6 = 4.2$

The largest square has sides of length 4.2 meters, so its area is the square of this, or $A = 4.2^{2} = 17.64$ square meters.

Of the choices, 18 square meters is closest.

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Example Question #21 : Squares

The areas of six squares form an arithmetic sequence. The smallest square has perimeter 16; the second smallest square has perimeter 20. Give the area of the largest of the six squares.

Possible Answers:

100

Correct answer:

Explanation:

The two smallest squares have perimeters 16 and 20, so their sidelengths are one fourth of these, or, respectively, 4 and 5. Their areas are the squares of these, or, respectively, 16 and 25. Therefore, in the arithmetic sequence,

$A_{1} = 16$

$A_{2} = 25$

and the common difference is d = 25 - 16 = 9 .

The area of the th smallest square is

$A_{n}= A_{1}+ (n-1)d$

Setting $A_{1} = 16, n = 6, d=9$ , the area of the largest (or sixth-smallest) square is

$A_{n}= 16+ (6-1)9 = 16 + 5 \cdot 9 = 16 + 45 = 61$

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Example Question #21 : Squares

Which is the greater quantity?

(a) The area of a square with sides of length meters

(b) The area of a square with perimeter 160 X centimeters

Possible Answers:

It cannot be determined which of (a) and (b) is greater

(a) and (b) are equal

(b) is the greater quantity

(a) is the greater quantity

Correct answer:

(a) is the greater quantity

Explanation:

A square with perimeter 160 X centimeters has sides of length one-fourth of this, or $160 X \div 4 = 40 X$ centimeters. Since one meter is equal to 100 centimeters, divide by 100 to get the equivalent in meters - this is

$40 X \div 100 = 0.4X$

meters.

The square in (b) has sidelength less than that of the square in (a), so its area is also less than that in (a).

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Example Question #23 : Squares

On the coordinate plane, Square A has as one side a segment with its endpoints at the origin and at the point with coordinates (A, B) . Square B has as one side a segment with its endpoints at the origin and at the point with coordinates (-B, -A) . and are both positive numbers and B > A . Which is the greater quantity?

(a) The area of Square A

(b) The area of Square B

Possible Answers:

(a) and (b) are equal

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

The length of a segment with endpoints (0,0) and (A, B) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = A$ , $y_{2} = B$ :

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

$d = \sqrt{(A-0)^{2}+(B-0)^{2}}$

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

The length of a segment with endpoints (0,0) and (-B, -A) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = -B$ , $y_{2} = -A$ :

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

$d = \sqrt{(-B-0)^{2}+(-A-0)^{2}}$

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

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

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

The sides are of equal length, so the squares have equal area. Note that the fact that B > A is irrelevant to the question.

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Example Question #231 : Geometry

On the coordinate plane, Square A has as one side a segment with its endpoints at the origin and at the point with coordinates (A, 2B) . Square B has as one side a segment with its endpoints at the origin and at the point with coordinates (2A, B) . and are both positive numbers and A > B . Which is the greater quantity?

(a) The area of Square A

(b) The area of Square B

Possible Answers:

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

The length of a segment with endpoints (0,0) and (A, 2B) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = A$ , $y_{2} = 2B$ :

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

$d = \sqrt{(A-0)^{2}+(2B-0)^{2}}$

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

$d = \sqrt{A ^{2}+4B^{2}}$

This is the length of one side of Square A; the area of the square is the square of this, or $A ^{2}+4B^{2}$ .

By similar reasoning, the length of a segment with endpoints (0,0) and (2A, B) is

$d = \sqrt{4A ^{2}+B^{2}}$

and the area of Square B is

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

Since A > B , and both are positive, it follows that

$A ^{2} > B ^{2}$

$3 A ^{2} >3 B ^{2}$

$3 A ^{2} +A ^{2} + B ^{2} >3 B ^{2} +A ^{2} + B ^{2}$

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

Square B has the greater area.

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Example Question #232 : Geometry

On the coordinate plane, Square A has as one side a segment with its endpoints at the origin and at the point with coordinates (A, B) . Square B has as one side a segment with its endpoints at the origin and at the point with coordinates (A+1, B-1) . and are both positive numbers. Which is the greater quantity?

(a) The area of Square A

(b) The area of Square B

Possible Answers:

(a) is the greater quantity

(a) and (b) are equal

(b) is the greater quantity

It is impossible to determine which is greater from the information given

Correct answer:

It is impossible to determine which is greater from the information given

Explanation:

It can be proved that the given information is insufficient to answer the question by looking at two cases.

Case 1: A = 4, B = 6

Square A has as a side a segment with endpoints at (0,0) and (4, 6) , the length of which can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = 4$ , $y_{2} = 6$ :

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

$d = \sqrt{(4-0)^{2}+(6-0 )^{2}}$

$d = \sqrt{4 ^{2}+6^{2}}$

$d = \sqrt{16+36}$

$d = \sqrt{52}$

This is the length of one side of Square A; the area of the square is the square of this, or 52.

Square B has as a side a segment with endpoints at (0,0) and (5,5) , the length of which can be found the same way:

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

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

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

$d = \sqrt{25+25}$

$d = \sqrt{50 }$

This is the length of one side of Square B; the area of the square is the square of this, or 50. This makes Square A the greater in area.

Case 2: A =B = 5

Square A has as a side a segment with endpoints at (0,0) and (5,5) ; this was found earlier to be a square of area 50.

Square B has as a side a segment with endpoints at (0,0) and (6, 4) , the length of which can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = 6$ , $y_{2} = 4$ :

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

$d = \sqrt{(6-0)^{2}+(4-0 )^{2}}$

$d = \sqrt{6 ^{2}+4 ^{2}}$

$d = \sqrt{36+16}$

$d = \sqrt{52}$

This is the length of one side of Square B; the area of the square is the square of this, or 52. This makes Square B the greater in area.

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Example Question #22 : Squares

Which is the greater quantity?

(a) The length of a diagonal of a square with sidelength 10 inches

(b) The hypotenuse of an isosceles right triangle with legs 10 inches each

Possible Answers:

It is impossible to tell which is greater from the information given.

(b) is greater.

(a) is greater.

(a) and (b) are equal.

Correct answer:

(a) and (b) are equal.

Explanation:

A diagonal of a square cuts the square into two isosceles right triangles, of which the diagonal is the common hypotenuse. Therefore, each figure is the hypotenuse of an isosceles right triangle with legs 10 inches, making them equal in length.

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

Track

The track at Peter Stuyvesant High School is a perfect square, as seen above, with sides of length 600 feet and a diagonal connecting two of the corners.

Les begins at Point A, takes the diagonal path directly to Point B, then runs counterclockwise around the square track twice. He then takes the diagonal from Point B back to Point A. Which of the following is closest to the distance he runs?

A hint: $\sqrt{2} \approx 1.414$

Possible Answers:

$1\frac{1}{2} \textup{ mi}$

$1 \textup{ mi}$

$1\frac{3}{4} \textup{ mi}$

$2 \textup{ mi}$

$1\frac{1}{4} \textup{ mi}$

Correct answer:

$1\frac{1}{4} \textup{ mi}$

Explanation:

The diagonal of a square has length $\sqrt{2}$ , or about 1.414, times the length of a side, which here is 600 feet; this makes the diagonal path about

$600 \times 1.414 \approx 848$ feet long.

Les runs around the square track twice, meaning that he runs the length of one side eight times; he also runs the length of the diagonal twice, This is a total of about

$600 \times 8 + 848 \times 2 = 4,800 + 1,696 = 6,496$ feet.

Divide by 5,280 to convert to miles:

$6,496 \div 5,280 \approx 1.23$

Of the given responses, $1\frac{1}{4}$ miles comes closest to the correct distance.

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Example Question #3 : How To Find The Length Of The Diagonal Of A Square

Track

The track at Franklin Pierce High School is a perfect square, as seen above, with sides of length 700 feet and a diagonal path connecting Points A and C.

Ellen wants to run three miles. Her plan is to begin at Point A, run along the diagonal path, run clockwise around the square track once, run along the diagonal path, run clockwise around the square track once, then repeat this pattern until she has run three miles. Where will she be when she is done?

A hint: $\sqrt{2} \approx 1.414$

Possible Answers:

On the square path between Point C and Point D

On the diagonal path between Point A and Point C

On the square path between Point A and Point B

On the square path between Point B and Point C

On the square path between Point D and Point A

Correct answer:

On the diagonal path between Point A and Point C

Explanation:

The diagonal of a square has length $\sqrt{2}$ , or about 1.414, times the length of a side, which here is 700 feet; this makes the diagonal path about

$700 \times 1.414 \approx 990$ feet long.

We will call one complete circuit one running of the diagonal, which is 990 feet long, and one running around the square; the completion of one complete circuit amounts to running a distance of

$990 + 4 \times 700 = 990 + 2,800 = 3,790$ feet.

Ellen seeks to run three miles, or

$5,280 \times 3 = 15, 840$ feet, which, divided by 3,790 feet, is about:

$15, 840 \div 3,790 \approx 4.17$ ,

or four complete circuits and 0.17 of a fifth.

After four complete circuits, Ellen is backat Point A. She has yet to run

$3,790 \times 0.17 = 629$ feet.

She will now run along the diagonal from Point A to Point C, but since the diagonal has length 990 feet, which is greater than 629 feet, she will finish running three miles when she is on this diagonal path.

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Example Question #4 : How To Find The Length Of The Diagonal Of A Square

Track

The track at Grant High School is a perfect square, as seen above, with sides of length 600 feet and a diagonal path connecting two of the corners.

Kenny begins at Point A, runs the path to Point C, and proceeds to run counterclockwise around the square track one complete time. He then runs again along the diagonal path from Point C to Point A.

Which is the greater quantity?

(a) The length of Kenny's run

(b) One mile

A hint: $\sqrt{2} \approx 1.414$

Possible Answers:

It is impossible to tell which is greater from the information given

(b) is greater

(a) and (b) are equal

(a) is greater

Correct answer:

(b) is greater

Explanation:

The diagonal of a square has length $\sqrt{2}$ , or about 1.414, times the length of a side, which here is 600 feet; this makes the diagonal path about

$600 \times 1.414 \approx 848$ feet long.

Kenny runs along this path twice, and he runs along the entire perimeter of the square path, so his run is about

$848 \times 2 + 600 \times 4 = 1,696 +2,400 = 4,096$ feet. Since one mile is equal to 5,280 feet, the greater quantity is (b).

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Squares

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #221 : Plane Geometry

Example Question #21 : Squares

Example Question #21 : Squares

Example Question #23 : Squares

Example Question #231 : Geometry

Example Question #232 : Geometry

Example Question #22 : Squares

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

Example Question #3 : How To Find The Length Of The Diagonal Of A Square

Example Question #4 : How To Find The Length Of The Diagonal Of A Square

All ISEE Upper Level Quantitative Resources

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