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ISEE Upper Level Quantitative : Quadrilaterals

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Example Questions

Example Question #1 : Quadrilaterals

Three of the interior angles of a quadrilateral measure $100 ^{\circ }$ , $105^{\circ }$ , and $110 ^{\circ }$ . What is the measure of the fourth interior angle?

Possible Answers:

$15 ^{\circ }$

$25^{\circ }$

$35^{\circ }$

$45^{\circ }$

This quadrilateral cannot exist.

Correct answer:

$45^{\circ }$

Explanation:

The measures of the angles of a quadrilateral have sum $360^{\circ }$ . If is the measure of the unknown angle, then:

x + 100 + 105 + 110 = 360

x + 315= 360

x + 315-315= 360-315

x = 45

The measure of the fourth angle is $45^{\circ }$ .

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Example Question #2 : Quadrilaterals

In a certain quadrilateral, three of the angles are $65^{\circ}$ , $120^{\circ}$ , and $34^{\circ}$ . What is the measure of the fourth angle?

Possible Answers:

$29^{\circ}$

$219^{\circ}$

$141^{\circ}$

$139^{\circ}$

$41^{\circ}$

Correct answer:

$141^{\circ}$

Explanation:

A quadrilateral has four angles totalling $360^{\circ}$ . So, first add up the three angles given. The sum is $219^{\circ}$ . Then, subtract that from 360. This gives you the missing angle, which is $141^{\circ}$ .

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Example Question #1 : Other Quadrilaterals

The angles of Quadrilateral A measure $80^{\circ }, 80^{\circ }, 80^{\circ }, x^{\circ }$

The angles of Pentagon B measure $100^{\circ }, 100^{\circ }, 100^{\circ }, 100^{\circ }, y^{\circ }$

Which is the greater quantity?

(A)

(B)

Possible Answers:

(A) is greater

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

(B) is greater

(A) and (B) are equal

Correct answer:

(B) is greater

Explanation:

The sum of the measures of the angles of a quadrilateral is $360^{\circ }$ ; the sum of the measures if a pentagon is $540^{\circ }$ . Therefore,

x + 80 + 80 + 80 = 360

x + 240 = 360

x = 120

and

y + 100+ 100+ 100+ 100 = 540

y + 4 00 = 540

y = 140

y > x , so (B) is greater.

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Example Question #3 : Quadrilaterals

Three of the interior angles of a quadrilateral measure $100 ^{\circ }$ , $105^{\circ }$ , and $110 ^{\circ }$ . What is the measure of the fourth interior angle?

Possible Answers:

$45^{\circ }$

This quadrilateral cannot exist.

$25^{\circ }$

$15 ^{\circ }$

$35^{\circ }$

Correct answer:

$45^{\circ }$

Explanation:

The measures of the angles of a quadrilateral have sum $360^{\circ }$ . If is the measure of the unknown angle, then:

x + 100 + 105 + 110 = 360

x + 315= 360

x + 315-315= 360-315

x = 45

The measure of the fourth angle is $45^{\circ }$ .

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Example Question #3 : Quadrilaterals

Which is the greater quantity?

(a) The perimeter of a square with sidelength 1 meter

(b) The perimeter of a regular pentagon with sidelength 75 centimeters

Possible Answers:

It is impossible to tell from the information given.

(b) is greater.

(a) and (b) are equal.

(a) is greater.

Correct answer:

(a) is greater.

Explanation:

(a) One meter is equal to 100 centimeters; a square with this sidelength has perimeter $100 \times 4 = 400$ centimeters.

(b) A regular pentagon has five congruent sides; since the sidelength is 75 centimeters, the perimeter is $75 \times 5 = 375$ centimeters.

This makes (a) greater.

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Example Question #2 : Squares

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

Which is the greater quantity?

(a) Twice the perimeter of Square 1

(b) The perimeter of Square 2

Possible Answers:

(a) and (b) are equal.

(a) is greater.

It is impossible to tell from the information given.

(b) is greater.

Correct answer:

(a) is greater.

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.

Since the perimeter of a square is four times its sidelength, Square 1 has perimeter ; Square 2 has perimeter $4 s \sqrt{2}$ , which is $\sqrt{2}$ times the perimeter of Square 1. $\sqrt{2} < 2$ , making the perimeter of Square 2 less than twice than the perimeter of Square 1.

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Example Question #1 : Squares

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 perimeters

(B) The median of their perimeters

Possible Answers:

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

(A) is greater

(A) and (B) are equal

(B) is greater

Correct answer:

(A) and (B) are equal

Explanation:

The perimeters of the squares are

$4 \times 1 = 4$ feet

$4 \times 2 = 8$ feet

$4 \times 3 = 12$ feet

$4 \times 4 = 16$ feet

$4 \times 5 = 20$ feet

The mean of the perimeters is their sum divided by five;

$(4+8+12+16+20) \div 5 = 60 \div 5 = 12$ feet.

The median of the perimeters is the value in the middle when they are arranged in ascending order; this can be seen to also be 12 feet.

The quantities are equal.

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Example Question #4 : Quadrilaterals

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

(A) The mean of their perimeters

(B) The median of their perimeters

Possible Answers:

(A) and (B) are equal

(A) is greater

(B) is greater

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

Correct answer:

(A) is greater

Explanation:

First find the perimeters of the squares:

$4 \times 100 = 400$ centimeters (one meter being 100 centimeters)

$4 \times 100 = 400$ centimeters

$4 \times 120 = 480$ centimeters

$4 \times 140 = 560$ centimeters

The mean of the perimeters is their sum divided by four:

$(400+400+480+560) \div 4 = 1,840 \div 4 = 460$ feet.

The median of the perimeters is the mean of the two values in the middle, assuming the values are in numerical order:

$(400 + 480) \div 2 = 880 \div 2 = 440$

The mean, (A), is greater.

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Example Question #1 : Quadrilaterals

The area of a square is $36t ^{2} + 84t + 49$ .

Give the perimeter of the square.

Possible Answers:

|42t|

|6t+7|

|36t+49|

|26t|

|24t+28|

Correct answer:

|24t+28|

Explanation:

The length of one side of a square is the square root of its area. The polynomial representing the area of the square can be recognized as a perfect square trinomial:

$36t ^{2} + 84t + 49$

$=( 6t )^{2} + 2\cdot 6t \cdot 7 + 7^{2}$

$= (6t+7)^{2}$

Therefore, the square root of the area is

$\sqrt{36t ^{2} + 84t + 49} = \sqrt{ (6t+7)^{2}} = |6t+7|$ ,

which is the length of one side.

The perimeter of the square is four times this length, or

4 |6t+7| = |24t+28| .

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Example Question #1 : Squares

The perimeters of six squares form an arithmetic sequence. The second-smallest square has sides that are two inches longer than those of the smallest square.

Which, if either, is the greater quantity?

(a) The perimeter of the third-smallest square

(b) The length of one side of the largest square

Possible Answers:

(a) and (b) are equal

(a) is greater

(b) is greater

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

Correct answer:

(a) is greater

Explanation:

Let the length of one side of the first square be $s_{1}$ . Then the length of one side of the second-smallest square is $s_{2}= s_{1}+ 2$ , and the perimeters of the squares are

$P_{1}= 4s_{1}$

and

$P_{2} = 4 s_{2}= 4 (s_{1}+ 2) = 4s_{1} + 8 = P_{1}+ 8$

This makes the common difference of the perimeters 8 units.

The perimeters of the squares being in arithmetic progression, the perimeter of the th-smallest square is

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

Since d = 8 , this becomes

$P_{n} = P_{1} + 8 (n-1)$

The perimeter of the third-smallest square is

$P_{3} = P_{1} + 8 (3-1) = P_{1}+ 8 \cdot 2 = P_{1} + 16$

The perimeter of the largest, or sixth-smallest, square is

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

The length of one side of this square is one fourth of this, or

$s_{6} = \frac{ P_{6} }{4}= \frac{ P_{1} + 40}{4} = \frac{1}{4} P_{1} + 10$

Therefore, we are comparing $P_{1} + 16$ and $\frac{1}{4} P_{1} + 10$ .

Since a perimeter must be positive,

$\frac{1}{4} P_{1} < P_{1}$ ;

also, 10 < 16 .

Therefore, regardless of the value of $P_{1}$ ,

$P_{1} + 16 > \frac{1}{4} P_{1} + 10$ ,

and

$P_{3}> s_{6}$ ,

making (a) the greater quantity.

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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 #1 : Quadrilaterals

Example Question #2 : Quadrilaterals

Example Question #1 : Other Quadrilaterals

Example Question #3 : Quadrilaterals

Example Question #3 : Quadrilaterals

Example Question #2 : Squares

Example Question #1 : Squares

Example Question #4 : Quadrilaterals

Example Question #1 : Quadrilaterals

Example Question #1 : Squares

All ISEE Upper Level Quantitative Resources

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