Acute / Obtuse Triangles - ISEE Upper Level Quantitative Reasoning

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Question

What is the slope of a line that passes through points and ?

Answer

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

Thus, if and , then:

$m=\frac{25-10}{10-5}=\frac{15}{5}=3$

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Question

What is the slope of a line that passes through points and ?

Answer

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

Thus, if and , then:

$m=\frac{6.5-2.5}{3.5-4.5}=\frac{4}{-1}=-4$

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Question

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) $30 ^{\circ }$

Answer

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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Question

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)

(b) 240

Answer

The angles of a hexagon measure a total of . From the information, we know that:

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

The quantities are equal.

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Question

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)

(b)

Answer

The angles of a hexagon measure a total of . From the information, we know that:

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

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

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

This makes (b) greater.

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Question

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)

Answer

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

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

, so (A) is greater.

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Question

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)

(B)

Answer

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

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

, so (B) is greater.

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Question

Seven angles of a convex octagon are congruent; the measure of the eighth is twice that of any one of the other seven. What is the measure of that eighth angle?

Answer

Let be the measure of any one of the seven congruent angles. Then the one non-congruent angle measures , and the sum of the angle measures in terms of is .

The angle measures of any convex octagon must add up to $180 (8-2) =1,080^{\circ}$ . So, to determine :

Therefore, the largest angle must measure $240^{\circ }$ , which is impossible since the measure of an angle cannot exceed $180 ^{\circ }$ .

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Question

Which is the greater quantity?

(a) The sum of the measures of the exterior angles of a thirty-sided polygon, one per vertex

(b) The sum of the measures of the exterior angles of a forty-sided polygon, one per vertex

Answer

The Polygon Exterior-Angle Theorem states that the sum of the measures of the exterior angles of any polygon, one per vertex, is $360^{\circ }$ . This makes both quantities equal.

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Question

Which is the greater quantity?

(a) The measure of an interior angle of an equilateral triangle

(b) The measure of an exterior angle of a regular octagon

Answer

Each angle of an equilateral triangle measures $60^{\circ}$ .

The sum of the exterior angles of any polygon, one per vertex, is $360^{\circ}$ . A regular octagon has eight sides, and, therefore, eight vertices; the measure of one exterior angle is $(360 \div 8 )^{\circ} = 45^{\circ}$ .

This makes (a) greater.

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Question

A regular polygon has exterior angles that measure $18^{\circ }$ each. Which is the greater quantity?

(a) The number of sides of this polygon

(b) 16

Answer

A regular polygon with 16 sides has exterior angles measuring

$\left ( \frac{360}{16} \right ) ^{\circ } = 22 .5 ^{\circ }$

The polygon in (a) has exterior angles that are narrower, so it must have more than 16 sides. (a) is greater.

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Question

A regular polygon has interior angles that measure $165^{\circ }$ each. Which is the greater quantity?

(a) The number of sides of the polygon

(b) 24

Answer

A regular polygon with 24 sides has interior angles measuring

$\left [\frac{180 \left ( 24-2\right ) }{24} \right ]^{\circ } = \left( \frac{180 \times 22 }{24} \right )^{\circ } = 165 ^{\circ }$

Therefore, the polygon in (a) has 24 sides, and the quantities are equal.

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Question

Polygon B is regular. The measure of each exterior angle of Polygon B is $30 ^{\circ }$ . Which of the following is the greater quantity?

(a) The number of sides of Polygon B

(b) 12

Answer

In any polygon, the sum of the measures of the exterior angles, one per vertex, is $360 ^{\circ }$ ; if the polygon is regular, its exterior angles have the same measure. If the polygon has sides - and vertices - then

$30^{\circ } \times N = 360 ^{\circ }$

and

$N = 360 ^{\circ } \div 30^{\circ } = 12$

This means the polygon has 12 sides, making the quantities equal.

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Question

Polygon A is a regular polygon with interior angles of measure $140^{\circ }$ .

Which is the greater quantity?

(a) The number of sides of Polygon A

(b) 10

Answer

If a regular polygon - one with congruent sides and congruent angles - has interior angles of measure $140^{\circ }$ , then its exterior angles each have measure $40^{\circ }$ . The sum of the measures of the exterior angles, one per vertex, is $360 ^{\circ }$ , so if the polygon has sides - and vertices - then

$40^{\circ } \times N = 360 ^{\circ }$

and

$N = 360 ^{\circ } \div 40^{\circ } = 9$

This means the polygon has 9 sides, making (b) the greater quantity.

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Question

A pentagon has five angles whose measures are $x^{\circ}, x^{\circ}, x^{\circ}, y^{\circ}, 2y^{\circ}$ .

Which quantity is greater?

(a)

(b) 180

Answer

The angles of a pentagon measure a total of . From the information, we know that:

$3\left ( x + y \right )= 540$

$3\left ( x + y \right )\div 3= 540\div 3$

making the two quantities equal.

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Question

Pentagon and hexagon are both regular, with their sidelengths equal. Diagonals $\overline{AC}$ and $\overline{UW }$ are constructed.

Which is the greater quantity?

(a)

(b)

Answer

Each diagonal, along with two consecutive sides of its polygon, forms a triangle. All of the sides of the pentagon and the hexagon are congruent to one another, so between the two triangles, there are two pairs of two congruent corresponding sides:

$\overline{AB} \cong \overline{ UV }$

$\overline{BC} \cong \overline{ VW }$

Their included angles, $\angle B$ and $\angle V$ , are interior angles of the pentagon and hexagon, respectively. The angle with greater measure will be opposite the longer side. We can use the Interior Angles Theorem to calculate the measures:

$m\angle B =\frac{ 180 (5-2)}{5} =108 ^{\circ }$

$m\angle V =\frac{ 180 (6-2)}{6} =120 ^{\circ }$

$m\angle V > m\angle B \Rightarrow UW > AC$

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Question

Pentagon and hexagon are both regular and have equal sidelengths. Diagonals $\overline{AC}$ and $\overline{UW }$ are constructed.

Which is the greater quantity?

(a) $m \angle BAC$

(b) $m \angle VUW$

Answer

In both situations, the two adjacent sides and the diagonal form an isosceles triangle.

By the Isosceles Triangle Theorem, $m \angle BAC = m \angle BCA$ and $m \angle VUW = m \angle VWU$ . Also, since the measures of the angles of a triangle total $180^{\circ }$ , we know that

$m \angle BAC + m \angle BCA + m \angle ABC = 180$

and

$m \angle VWU + m \angle VUW + m \angle UVW = 180$ .

We can use these equations to compare $m \angle BAC$ and $m \angle VUW$ .

(a) $m \angle ABC = \frac{ 180 (5-2)}{5} = 108 ^{\circ }$

$m \angle BAC + m \angle BCA + m \angle ABC = 180 ^{\circ }$

$m \angle BAC + m \angle BAC + 108 ^{\circ } = 180 ^{\circ }$

$2\cdot m \angle BAC + 108 ^{\circ }= 180 ^{\circ }$

$2\cdot m \angle BAC + 108 ^{\circ } -108 ^{\circ } = 180 ^{\circ }-108 ^{\circ }$

$2\cdot m \angle BAC = 72 ^{\circ }$

$2 \cdot m \angle BAC \div 2 = 72 ^{\circ } \div 2$

$m \angle BAC = 36 ^{\circ }$

(b) $m \angle UVW = \frac{ 180 (6-2)}{6} = 120 ^{\circ }$

$m \angle VWU + m \angle VUW + m \angle UVW = 180 ^{\circ }$

$m \angle VUW + m \angle VUW + 120^{\circ } = 180 ^{\circ }$

$2 \cdot m \angle VUW + 120^{\circ } = 180 ^{\circ }$

$2 \cdot m \angle VUW + 120^{\circ } -120^{\circ } = 180 ^{\circ }-120^{\circ }$

$2 \cdot m \angle VUW = 60^{\circ }$

$2 \cdot m \angle VUW \div 2 = 60^{\circ }\div 2$

$m \angle VUW = 30^{\circ }$

$m \angle BAC > m \angle VUW$

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Question

You are given pentagon .

$\angle A \cong \angle B \cong \angle C$

$\angle D \cong \angle E$

Which is the greater quantity?

(A) $108 ^{\circ }$

(B) $m\angle D$

Answer

It is impossible to tell, as scenarios can be constructed that would allow $m\angle D$ to be less than, equal to, or greater than 108, keeping in mind that the sum of the degree measures of a pentagon is .

Case 1: The pentagon is regular, so all five angles are of the same measure:

$540 \div 5 = 108$

This fits the conditions of the problem and makes the two quantities equal.

Case 2: $m \angle A = m \angle B =m \angle C = 100; m \angle D = m \angle E = 120$

The sum of the angle measures is therefore

This also fits the conditions of the problem, and makes (B) greater.

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Question

In Pentagon ,

$m \angle A = 80^{\circ }$

The other four angles are congruent to one another.

What is $m \angle B$ ?

Answer

The degree measures of a pentagon, which has five angles, total .

$m \angle A = 80^{\circ }$ .

Let $x = m \angle B$ . Then since the other three angles all have the same measure as $\angle B$ ,

$x = m \angle B = m \angle C = m \angle D = m \angle E$

Therefore, we can set up, and solve for in, the equation

$4x \div 4 = 460 \div 4$

$m \angle B = 115^{\circ }$

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Question

Note: Figure NOT drawn to scale

In the above figure, $\overline{AB}$ and $\overline{BC}$ are adjacent sides of a regular pentagon; $\overline{AB}$ and $\overline{BD}$ are adjacent sides of a regular hexagon. Which of the following is the greater quantity?

(a) $m \angle CBD$

(b) $m \angle CBA$

Answer

Extend $\overline{AB}$ as seen below:

$\angle CBA$ , as an interior angle of a regular pentagon (five-sided polygon), has measure

$m \angle CBA =\left ( \frac{180(5-2)}{5} \right )^{\circ } = \left ( \frac{180(3)}{5} \right )^{\circ } = 108^{\circ }$ .

Its exterior angle $\angle CBE$ has measure $(180- 108)^{\circ } = 72^{\circ }$ .

$\angle DBA$ , as an interior angle of a regular hexagon (six-sided polygon), has measure

$m \angle DBA =\left ( \frac{180(6-2)}{6} \right )^{\circ } = \left ( \frac{180(4)}{6} \right )^{\circ } = 120^{\circ }$ .

Its exterior angle $\angle DBE$ has measure $(180- 120)^{\circ } = 60^{\circ }$ .

Add the measures of $\angle CBE$ and $\angle DBE$ to get that of $\angle CBD$ :

$m \angle CBD = m \angle CBE + m \angle DBE = 72^{\circ }+ 60 ^{\circ } = 132^{\circ }$ .

$m \angle CBD > m \angle CBA$ .

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