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ISEE Upper Level Quantitative : How to find the area of a square

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Example Question #11 : Squares

A diagonal of a square has length $2^{x}$ . Give its area.

Possible Answers:

$2^{2x}$

$2^{2x- \frac{1}{2}}$

$2^{2x+ \frac{1}{2}}$

$2^{2x+1}$

$2^{2x-1}$

Correct answer:

$2^{2x-1}$

Explanation:

A square being a rhombus, its area can be determined by taking half the product of the lengths of its (congruent) diagonals:

$A = \frac{1}{2} \cdot 2^{x} \cdot 2^{x} = 2 ^{-1} \cdot 2^{x} \cdot 2^{x} = 2^{-1 + x + x } = 2^{2x-1}$

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Example Question #21 : Squares

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:

24 square meters

20 square meters

22 square meters

18 square meters

16 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 #22 : 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 #221 : Plane Geometry

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:

(a) and (b) are equal

(b) is the greater quantity

(a) is the greater quantity

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

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 #11 : How To Find The Area Of A Square

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

(b) is the greater quantity

(a) is the greater quantity

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

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 : Plane 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:

(a) is the greater quantity

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

(b) is the greater quantity

(a) and (b) are equal

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 #21 : 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 (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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ISEE Upper Level Quantitative : How to find the area of a square

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #11 : Squares

Example Question #21 : Squares

Example Question #22 : Squares

Example Question #221 : Plane Geometry

Example Question #11 : How To Find The Area Of A Square

Example Question #231 : Plane Geometry

Example Question #21 : Squares

All ISEE Upper Level Quantitative Resources

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