How to find the area of a square - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

The perimeter of a square is one yard. Which is the greater quantity?

(a) The area of the square

(b) $\frac{1}{2}$ square foot

Answer

One yard is equal to three feet, so the length of one side of a square with this perimeter is $\frac{3}{4}$ feet. The area of the square is $\frac{3}{4}\times \frac{3}{4} = \frac{9}{16}$ square feet. $\frac{9}{16} > \frac{1}{2}$ , making (a) greater.

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Question

Square 1 is inscribed inside a circle. The circle is inscribed inside Square 2.

Which is the greater quantity?

(a) Twice the area of Square 1

(b) The area of Square 2

Answer

Let be the sidelength of Square 1. Then the length of a diagonal of this square - which is $\sqrt{2}$ times this sidelength, or $s \sqrt{2}$ , by the $45^{\circ }-45^{\circ }-90^{\circ }$ Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2.

Therefore, Square 1 has area $A = s^{2}$ , and Square 2 has area $A = \left ( s \sqrt{2 } \right )^{2} = 2s^{2}$ , twice that of Square 1.

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Question

Which is the greater quantity?

(A) The area of a square with sidelength one foot

(B) The area of a rectangle with length nine inches and height fourteen inches

Answer

The area of a square is the square of its sidelength, which here is 12 inches:

$A = 12^{2} = 144$ square inches.

The area of a rectangle is its length multiplied by its height, which, respectively, are 9 inches and 14 inches:

$A = 9 \cdot 14 = 126$ square inches.

The square has the greater area.

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Question

A square lawn has sidelength twenty yards. Give its area in square feet.

Answer

20 yards converts to $20 \times3 = 60$ feet. The area of a square is the square of its sidelength, so the area in square feet is $60^{2} = 60 \times 60 = 3,600$ square feet.

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Question

Five squares have sidelengths one foot, two feet, three feet, four feet, and five feet.

Which is the greater quantity?

(A) The mean of their areas

(B) The median of their areas

Answer

The areas of the squares are:

$1^{2} = 1$ square foot

$2^{2} = 4$ square feet

$3^{2} = 9$ square feet

$4^{2}= 16$ square feet

$5^{2}= 25$ square feet

Therefore, we are comparing the mean and the median of the set $\left { 1, 4, 9, 16, 25\right }$ .

The mean of this set is the sum divided by 5:

$(1 + 4 + 9 + 16 + 25) \div 5 = 55 \div 5 = 11$

The median is the middle element after arrangement in ascending order, which is 9.

This makes (A), the mean, greater.

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Question

Four squares have sidelengths one meter, 120 centimeters, 140 centimeters, and 140 centimeters. Which is the greater quantity?

(A) The mean of their areas

(B) The median of their areas

Answer

The areas of the squares are:

$100 \cdot 100 = 10,000$ square centimeters (one meter being 100 centimeters)

$120 \cdot 120 = 14,400$ square centimeters

$140 \cdot 140 = 19,600$ square centimeters

$140 \cdot 140 = 19,600$ square centimeters

The mean of these four areas is their sum divided by four:

$(10,000+14,400+19,600+19,600) \div 4$

$= 63,600 \div 4 =15,900$ square centimeters.

The median is the mean of the two middle values, or

$( 14,400+19,600 ) \div 2 = 34,000 \div 2 = 17,000$ square centimeters.

The median, (B), is greater.

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Question

Rectangle A and Square B both have perimeter 2 meters. Rectangle A has width 25 centimeters. The area of Rectangle A is what percent of the area of Square B?

Answer

The perimeter of a rectangle can be given by the formula

Rectangle A has perimeter 2 meters, which is equal to 200 centimeters, and width 25 centimeters, so the length is:

The dimensions of Rectangle A are 75 centimeters and 25 centimeters, so its area is

$75 \times 25 = 1,875$ square centimeters.

The sidelength of a square is one-fourth its perimeter, which here is

$\frac{200}{4} = 50$ centimeters; its area is therefore

$50 ^{2} = 2,500$ square centimeters.

The area of Rectangle A is therefore

$\frac{1,875}{2,500} \times 100 = 75 %$

that of Square B.

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Question

The sidelength of Square A is three-sevenths that of Square B. What is the ratio of the area of Square B to that of Square A?

Answer

Since the ratio is the same regardless of the sidelengths, then for simplicity's sake, assume the sidelength of Square B is 7. The area of Square B is therefore the square of this, or 49.

Then the sidelength of Square A is three-sevenths of 7, or 3. Its area is the square of 3, or 9.

The ratio of the area of Square B to that of Square A is therefore 49 to 9.

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Question

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.

Answer

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

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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Question

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.

Answer

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 .

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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Question

The perimeter of a square is . Give the area of the square in terms of .

Answer

The length of one side of a square is one fourth its perimeter. Since the perimeter of the square is , the length of one side is

$(12x+16) \div 4 = 3x+ 4$

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

$( 3x+ 4)^{2}= (3x)^{2}+ 2 \cdot 3x \cdot 4 + 4 ^{2} = 9x^{2}+24x + 16$

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Question

The sidelength of a square is $2^{x}$ . Give its area in terms of .

Answer

The area of a square is the square of its sidelength. Therefore, square $2^{x}$ :

$\left (2^{x} \right ) ^{2} = 2 ^{2 x}$

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Question

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

Answer

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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Question

Which is the greater quantity?

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

(b) The area of a square with perimeter centimeters

Answer

A square with perimeter 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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Question

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

(a) The area of Square A

(b) The area of Square B

Answer

The length of a segment with endpoints and 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 and 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 is irrelevant to the question.

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Question

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

(a) The area of Square A

(b) The area of Square B

Answer

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

Case 1:

Square A has as a side a segment with endpoints at and , 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 and , 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:

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

Square B has as a side a segment with endpoints at and , 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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Question

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

(a) The area of Square A

(b) The area of Square B

Answer

The length of a segment with endpoints and 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 and is

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

and the area of Square B is

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

Since , 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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