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ISEE Upper Level Quantitative : Quadrilaterals

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

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All ISEE Upper Level Quantitative Resources

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Example Questions

Example Question #2 : Squares

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

Possible Answers:

$2 ^{x+4}$

$2 ^{x+2}$

$2 ^{x+1}$

$2 ^{2 x}$

$2 ^{4 x}$

Correct answer:

$2 ^{x+2}$

Explanation:

The perimeter of a square is four times the length of a side, which here is $2 ^{x }$ :

$P = 4 \cdot 2^{x} = 2 ^{2} \cdot 2^{x} =2 ^{x+2}$

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Example Question #2 : Squares

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

Possible Answers:

$2 ^{x+\frac{3}{2}}$

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

$2^{2x}$

$2^{2x-1}$

$2^{3x}$

Correct answer:

$2 ^{x+\frac{3}{2}}$

Explanation:

The length of a side of a square can be determined by dividing the length of a diagonal by $\sqrt{2}$ - that is, $2 ^{\frac{1}{2}}$ . A diagonal has length $2^{x}$ , so the sidelength is

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

Multiply this by four to get the perimeter:

$4 \cdot 2 ^{x-\frac{1}{2}} = 2^{2} \cdot 2 ^{x-\frac{1}{2}} = 2 ^{x-\frac{1}{2}+2} = 2 ^{x+\frac{3}{2}}$

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Example Question #1 : Squares

The perimeters of six squares form an arithmetic sequence. The smallest square has area 9; the second smallest square has area 25. Give the perimeter of the largest of the six squares.

Possible Answers:

None of the other responses gives the correct answer.

Correct answer:

Explanation:

The two smallest squares have areas 9 and 25, so their sidelengths are the square roots of these, or, respectively, 3 and 5. Their perimeters are the sidelengths multiplied by four, or, respectively, 12 and 20. Therefore, in the arithmetic sequence,

$P_{1} = 12$

$P_{2} = 20$

and the common difference is d = 20- 12 = 8 .

The perimeter of the th smallest square is

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

Setting $P_{1} = 12, n = 6, d=8$ , the perimeter of the largest (or sixth-smallest) square is

$P_{6}=P_{1}+ (6-1)8 = 12+ 5 \cdot 8 = 12 + 40 = 52$ .

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Example Question #11 : Quadrilaterals

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

Possible Answers:

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell form the information given.

Correct answer:

(a) is greater.

Explanation:

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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Example Question #12 : Quadrilaterals

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

Possible Answers:

(b) is greater.

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Correct answer:

(a) and (b) are equal.

Explanation:

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

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

Possible Answers:

(A) and (B) are equal

(B) is greater

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

(A) is greater

Correct answer:

(A) is greater

Explanation:

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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Example Question #13 : Quadrilaterals

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

Possible Answers:

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

$3,200\textrm{ ft}^{2}$

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

$2,800\textrm{ ft}^{2}$

$4,200\textrm{ ft}^{2}$

Correct answer:

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

Explanation:

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

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?

Possible Answers:

$66 \frac{2}{3} \%$

$75 \%$

$83 \frac{1}{3} \%$

$80 \%$

$87 \frac{1}{2} \%$

Correct answer:

$75 \%$

Explanation:

The perimeter of a rectangle can be given by the formula

P = 2(L+W)

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

2(L+W) = P

2(L+25) = 200

2L + 50= 200

2L = 150

L = 75

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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Example Question #14 : Quadrilaterals

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?

Possible Answers:

$7\textrm{ to }3$

$49 \textrm{ to } 9$

$9\textrm{ to }7$

$14\textrm{ to }3$

None of the other answers give the correct ratio.

Correct answer:

$49 \textrm{ to } 9$

Explanation:

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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Example Question #7 : How To Find The Area Of A Square

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

Possible Answers:

(A) is greater

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

(A) and (B) are equal

(B) is greater

Correct answer:

(A) is greater

Explanation:

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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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Quadrilaterals

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #2 : Squares

Example Question #2 : Squares

Example Question #1 : Squares

Example Question #11 : Quadrilaterals

Example Question #12 : Quadrilaterals

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

Example Question #13 : Quadrilaterals

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

Example Question #14 : Quadrilaterals

Example Question #7 : How To Find The Area Of A Square

All ISEE Upper Level Quantitative Resources

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