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ISEE Upper Level Quantitative : Geometry

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 #91 : Geometry

What is the slope of a line that passes through points (4,12) and (8, 20) ?

Possible Answers:

2.5

0.5

Correct answer:

Explanation:

The equation for solving for the slope of a line is $m=\frac{y2-y1}{x2-x1}$ .

Thus, if (x1,y1)=(4,12) and (x2,y2)=(8,20) , then:

$m=\frac{20-12}{8-4}=\frac{8}{4}=2$

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Example Question #92 : Plane Geometry

Tangents 1

Figure NOT drawn to scale

In the above figure, $\overline{TN}$ and $\overline{TO}$ are tangent segments. The ratio of the length of $\overarc{NPO}$ to that of $\overarc{N O}$ is 5 to 3. Which is the greater quantity?

(a) $m \angle NTO$

(b) $30 ^{\circ }$

Possible Answers:

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

(a) and (b) are equal

(a) is the greater quantity

(b) is the greater quantity

Correct answer:

(a) is the greater quantity

Explanation:

For the sake of simplicity, let us assume that the lengths of $\overarc{NPO}$ and $\overarc{N O}$ are 5 and 3; this reasoning depends only on their ratio and not their actual length. The circumference of the circle is the sum of the lengths, which is 8, so $\overarc{NPO}$ and $\overarc{N O}$ comprise $\frac{5}{8}$ and $\frac{3}{8}$ of the circle, respectively. Therefore,

$m \overarc{NPO} = \frac{5}{8} \cdot 360 ^{\circ } = 225 ^{\circ }$ ; and

$m \overarc{N O} = \frac{3}{8} \cdot 360 ^{\circ } = 135 ^{\circ }$ .

If two tangents are drawn to a circle, the measure of the angle they form is half the difference of the measures of the arcs they intercept, so

$m \angle NTO = \frac{1}{2}(m \overarc{NPO} - m \overarc{N O} ) = \frac{1}{2}(225 ^{\circ }- 135^{\circ } ) = \frac{1}{2}(90^{\circ } ) = 45^{\circ }$

This is greater than $30 ^{\circ }$ .

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Example Question #91 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Which is the greater quantity?

(a) The length of the line segment connecting (0,0) and (a,b)

(b) The length of the line segment connecting (a,0) and (0,b)

Possible Answers:

(a) and (b) are equal.

It is impossible to tell from the information given.

(b) is greater.

(a) is greater.

Correct answer:

(a) and (b) are equal.

Explanation:

(a) The length of the line segment connecting (0,0) and (a,b) is

$\sqrt{(a -0)^{2}+(b -0)^{2} } = \sqrt{a^{2}+b^{2} }$ .

(b) The length of the line segment connecting (a,0) and (0,b) is

$\sqrt{(a -0)^{2}+(0-b)^{2} } = \sqrt{a^{2}+\left ( - b \right ) ^{2} }= \sqrt{a^{2}+b^{2} }$ .

The segments have equal length.

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Example Question #91 : Plane Geometry

Which is the greater quantity?

(a) The area of a regular hexagon with sidelength 1

(b) The area of an equilateral triangle with sidelength 2

Possible Answers:

It is impossible to tell from the information given

(b) is greater

(a) is greater

(a) and (b) are equal

Correct answer:

(a) is greater

Explanation:

A regular hexagon with sidelength s = 1 can be seen as a composite of six equilateral triangles, each with sidelength s = 1 . Since area is in direct proportion to the square of the sidelength, the area of the equilateral triangle with sidelength s = 2 is equal to that of four of those triangles. This makes the hexagon greater in area, and it makes (a) the greater quantity.

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Example Question #1 : Hexagons

Which is the greater quantity?

(a) The perimeter of a regular pentagon with sidelength 1 foot

(b) The perimeter of a regular hexagon with sidelength 10 inches

Possible Answers:

(b) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) is greater.

Correct answer:

(a) and (b) are equal.

Explanation:

The sides of a regular polygon are congruent, so in each case, multiply the sidelength by the number of sides to get the perimeter.

(a) Since one foot equals twelve inches, $5 \times 12 = 60$ inches.

(b) Multiply: $6 \times 10 = 60$ inches

The two polygons have the same perimeter.

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Example Question #92 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

A hexagon has six angles with measures $(x-5)^{\circ}, x^{\circ}, (x+5)^{\circ}, (y-10)^{\circ}, y^{\circ}, (y + 10)^{\circ}.$

Which quantity is greater?

(a) x + y

(b) 240

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:

The angles of a hexagon measure a total of 180 (6-2) = 720 . From the information, we know that:

(x - 5) + x + (x+5) + (y-10) + y + (y+10) = 720

x+ x + x + y+ y + y - 5+5 -10 +10= 720

3x + 3y = 720

3 (x + y) = 720

$3 (x + y) \div 3 = 720 \div 3$

x + y = 240

The quantities are equal.

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Example Question #93 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

A hexagon has six angles with measures $(x-10)^{\circ}, x^{\circ}, (x+5)^{\circ}, (y-10)^{\circ}, y^{\circ}, (y + 20)^{\circ}.$

Which quantity is greater?

(a) x + y

(b) 240

Possible Answers:

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

Correct answer:

(b) is greater.

Explanation:

The angles of a hexagon measure a total of 180 (6-2) = 720 . From the information, we know that:

(x-10)+x+(x+5)+ (y-10)+ y+ (y + 20)= 720

x + x + x + y + y + y -10+5 -10 + 20= 720

3x + 3y +5= 720

3x + 3y +5-5= 720-5

3x + 3y = 715

$3\left ( x + y \right )= 715$

$3\left ( x + y \right ) \div 3= 715\div 3$

$x + y = 238 \frac{1}{3} < 240$

This makes (b) greater.

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Example Question #94 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

The angles of Hexagon A measure

$A ^{\circ }, B ^{\circ }, 150 ^{\circ }, 150 ^{\circ }, 150 ^{\circ }, 150 ^{\circ }$

The angles of Octagon B measure

$C ^{\circ }, D^{\circ }, 150^{\circ },150^{\circ },150^{\circ },150^{\circ },150^{\circ },150^{\circ }$

Which is the greater quantity?

(A) A + B

(B) C + D

Possible Answers:

(B) is greater

(A) is greater

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

(A) and (B) are equal

Correct answer:

(B) is greater

Explanation:

The sum of the measures of a hexagon is $180 ^{\circ } \times (6-2)= 720 ^{\circ }$ . Therefore,

A + B + 150 + 150 + 150 + 150 = 720

A + B +600= 720

A + B = 120

The sum of the measures of an octagon is $180 ^{\circ } \times (8-2)= 1,080 ^{\circ }$ . Therefore,

C + D + 150 + 150 + 150 + 150+ 150 + 150 = 1,080

C + D + 900 = 1,080

C + D = 180

C + D > A + B , so (B) is greater.

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Example Question #5 : Hexagons

The angles of Pentagon A measure $110^{\circ }, 110^{\circ }, 110^{\circ }, y^{\circ }, y^{\circ }$

The angles of Hexagon B measure $130^{\circ }, 130^{\circ }, 130^{\circ }, 130^{\circ },z^{\circ }, z^{\circ }$

Which is the greater quantity?

(A)

(B)

Possible Answers:

(B) is greater

(A) and (B) are equal

(A) is greater

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

Correct answer:

(A) is greater

Explanation:

The sum of the measures of the angles of a pentagon is $180^{\circ } \times (5-2) =540^{\circ }$ . Therefore,

y + y + 110 + 110 + 110 = 540

2y +330 = 540

2y = 210

y = 105

The sum of the measures of a hexagon is $180^{\circ } \times (6-2) =720 ^{\circ }$ . Therefore,

z + z + 130 + 130 + 130 + 130 = 720

2z + 520 = 720

2z= 200

z = 100

y> z , so (A) is greater.

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Example Question #1 : How To Find The Length Of The Side Of A Hexagon

Right_triangle

A regular hexagon has the same perimeter as the above right triangle. What is the length of one side of the hexagon?

Possible Answers:

$12 \frac{2}{5}\textrm{ in}$

$22 \frac{2}{5} \textrm{ in}$

The length cannot be determined from the information given.

$18 \frac{2}{3} \textrm{ in}$

$10 \frac{1}{3}\textrm{ in}$

Correct answer:

$18 \frac{2}{3} \textrm{ in}$

Explanation:

By the Pythagorean Theorem, the hypotenuse of the right triangle is

$\sqrt{14^{2}+48^{2}} = \sqrt{196+2,304} = \sqrt{2,500} = 50$ inches, making its perimeter

14 + 48 + 50 =112 inches.

The regular hexagon, which has six sides of equal length, has the same perimeter, so each side measures

$112 \div 6 = 18 \frac{2}{3}$ inches.

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Geometry

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #91 : Geometry

Example Question #92 : Plane Geometry

Example Question #91 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #91 : Plane Geometry

Example Question #1 : Hexagons

Example Question #92 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #93 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #94 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #5 : Hexagons

Example Question #1 : How To Find The Length Of The Side Of A Hexagon

All ISEE Upper Level Quantitative Resources

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