Rhombuses - ISEE Upper Level Quantitative Reasoning

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Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

Answer

The two marked angles are same-side exterior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

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Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

Answer

The two marked angles are same-side interior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

$2y \div 2 = \left ( 184 -x \right )\div 2$

$y= 92 - \frac{1}{2}x$

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Question

Examine the above diagram. Which of the following statements must be true whether or not and are parallel?

Answer

Four statements can be eliminated by the various parallel theorems and postulates. Congruence of alternate interior angles or corresponding angles forces the lines to be parallel, so

$\angle 5 \cong \angle 4 \Rightarrow l \parallel m$ and

$\angle 2 \cong \angle 6 \Rightarrow l \parallel m$ .

Also, if same-side interior angles or same-side exterior angles are supplementary, the lines are parallel, so

$m \angle 4 + m \angle 7 = 180 ^{\circ } \Rightarrow l \parallel m$ and

$m \angle 1 + m \angle 6 = 180 ^{\circ } \Rightarrow l \parallel m$ .

However, $\angle 5 \cong \angle 8$ whether or not $l \parallel m$ since they are vertical angles, which are always congruent.

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Question

Examine the above diagram. What is ?

Answer

By angle addition,

$\left ( x - 13\right )+ 100 + (y + 25) = 180$

$x - 13\right+ 100 + y + 25 = 180$

$x + y - 13\right+ 100+ 25 = 180$

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Question

$\angle 1$ and $\angle 2$ are supplementary; $\angle 1$ and $\angle 3$ are complementary.

$m \angle 2 = 100 ^{\circ }$ .

What is $m \angle 3$ ?

Answer

Supplementary angles and complementary angles have measures totaling $180^{\circ }$ and $90^{\circ }$ , respectively.

$m \angle 2 = 100 ^{\circ }$ , so its supplement $\angle 1$ has measure

$m \angle 1 = 180^{\circ } - m \angle 2 = 180^{\circ } - 100 ^{\circ } = 80^{\circ }$

$\angle 3$ , the complement of $\angle 1$ , has measure

$m \angle 3 = 90^{\circ } - m \angle 1 = 90^{\circ } - 80 ^{\circ } = 10^{\circ }$

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Question

Note: Figure NOT drawn to scale.

In the above figure, $m \angle 1=\left ( x+17 \right )^{\circ}$ and $m \angle 2= \left (7x+11 \right )^{\circ}$ . Which of the following is equal to $m \angle 1$ ?

Answer

$\angle 1$ and $\angle 2$ form a linear pair, so their angle measures total $180^{\circ}$ . Set up and solve the following equation:

$m \angle 1+ m \angle 2 = 180^{\circ}$

$\left ( x+17 \right ) + \left (7x+11 \right ) = 180$

$m \angle 1=\left ( x+17 \right )^{\circ} = \left ( 19+17 \right )^{\circ} = 36^{\circ}$

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Question

Two angles which form a linear pair have measures $(2x+72)^{\circ}$ and $(5x-125)^{\circ}$ . Which is the lesser of the measures (or the common measure) of the two angles?

Answer

Two angles that form a linear pair are supplementary - that is, they have measures that total $180^{\circ}$ . Therefore, we set and solve for in this equation:

$x = \frac{233}{7}= 33 \frac{2}{7}$

The two angles have measure

$2 \cdot 33 \frac{2}{7} +72= 138 \frac{4}{7} ^{\circ}$

and

$5 \cdot 33 \frac{2}{7} -125 = 41\frac{3}{7}^{\circ}$

$41\frac{3}{7}^{\circ}$ is the lesser of the two measures and is the correct choice.

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Question

Two vertical angles have measures $(3x+14)^{\circ }$ and $\left ( 5x-17\right )^{\circ }$ . Which is the lesser of the measures (or the common measure) of the two angles?

Answer

Two vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure. Therefore, we set up and solve the equation

$x= 15\frac{1}{2}$

$3x+14 = 3 \cdot 15\frac{1}{2} + 14 = 46\frac{1}{2} + 14= 60\frac{1}{2} ^{\circ }$

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Question

A line intersects parallel lines and . $\angle 1$ and $\angle 2$ are corresponding angles; $\angle 1$ and $\angle 3$ are same side interior angles.

$m \angle 1 = \left (3x+2y \right )^{\circ }$

$m \angle 2 = \left ( 4x+21\right )^{\circ }$

$m \angle 3 = \left ( 2y-27\right )^{\circ}$

Evaluate .

Answer

When a transversal such as crosses two parallel lines, two corresponding angles - angles in the same relative position to their respective lines - are congruent. Therefore,

Two same-side interior angles are supplementary - that is, their angle measures total 180 - so

We can solve this system by the substitution method as follows:

Backsolve:

, which is the correct response.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the measure of $\angle 1$ .

Answer

The top and bottom angles, being vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure, so

,

or, simplified,

The right and bottom angles form a linear pair, so their degree measures total 180. That is,

Substitute for :

The left and right angles, being vertical angles, have the same measure, so, since the right angle measures $\left (8x \right )^{\circ } =\left (8 \cdot 15 \right )^{\circ } = 120 ^{\circ }$ , this is also the measure of the left angle, $\angle 1$ .

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Question

Figure NOT drawn to scale

The above figure shows Trapezoid , with $\overline{RT}$ and $\overline{RA }$ tangent to the circle. $m \angle T = 106 ^{\circ }$ ; evaluate $m \angle A$ .

Answer

By the Same-Side Interior Angle Theorem, since $\overline{TR} || \overline{AP}$ , $\angle T$ and $\angle P$ are supplementary - that is, their degree measures total $180 ^{\circ }$ . Therefore,

$m \angle P + m \angle T = 180 ^{\circ }$

$m \angle P = 180 ^{\circ } - m \angle T$

$m \angle P = 180 ^{\circ } - 106 ^{\circ } = 74^{\circ }$

$\angle P$ is an inscribed angle, so the arc it intercepts, $\overarc{TA}$ , has twice its degree measure;

$m \overarc{TA} = 2 \cdot m \angle P =2 \cdot 74^{\circ } = 148 ^{\circ }$ .

The corresponding major arc, $\overarc{TPA}$ , has as its measure

$m \overarc{TPA} = 360 ^{\circ }- m \overarc{TA} = 360 ^{\circ }- 148 ^{\circ }= 212 ^{\circ }$

The measure of an angle formed by two tangents to a circle is equal to half the difference of those of its intercepted arcs:

$m \angle R = \frac{1}{2} \left (m \overarc{TPA} - m \overarc{TA} \right )$

$m \angle R = \frac{1}{2} \left (212 ^{\circ } -148 ^{\circ } \right )= \frac{1}{2} \cdot 64 ^{\circ } = 32^{\circ }$

Again, by the Same-Side Interior Angles Theorem, $\angle R$ and $\angle A$ are supplementary, so

$m \angle A + m \angle R= 180 ^{\circ }$

$m \angle A = 180 ^{\circ } - m \angle R$

$m \angle A = 180 ^{\circ } - 32 ^{\circ } = 148^{\circ }$

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Question

Note: Figure NOT drawn to scale.

Which of the following is the greater quantity?

(A) The perimeter of the triangle

(B) 90

Answer

The longest side of the triangle appears opposite the angle of greatest measure. The side of length 30 appears opposite an angle of measure $180 ^{\circ } - (65 ^{\circ } + 65 ^{\circ } ) = 50 ^{\circ }$ . Therefore, the sides opposite the $65^{\circ }$ angles must have lengths greater than 30.

If we let this common length be , then

The perimeter of the triangle is therefore greater than 90.

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Question

A square, a regular pentagon, a regular hexagon, and a regular octagon have the same sidelength. Which is the greater quantity?

(A) The mean of their perimeters

(B) The median of their perimeters

Answer

The answer is independent of the sidelength, so we can assume without loss of generality that the sidelength is 1. The square, the pentagon, the hexagon, and the octagon have 4, 5, 6, and 8 sides of equal length, respectively, so their perimeters are 4, 5, 6, and 8. The mean of these four perimeters is

$\left ( 4+5+6+8\right ) \div 4 = 23 \div 4 = 5.75$ units.

The median is the mean of the middle two perimeters, which are 5 and 6:

$\left ( 5 + 6\right ) \div 2 = 5.5$

The mean, (A), is greater.

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Question

An equilateral triangle, a square, a regular pentagon, a regular hexagon, and a regular octagon have the same sidelength. Which is the greater quantity?

(A) The median of their perimeters

(B) The midrange of their perimeters

Answer

The answer is independent of the sidelength, so we can assume without loss of generality that the sidelength is 1. The equilateral triangle, the square, the pentagon, the hexagon, and the octagon have 3, 4, 5, 6, and 8 sides of equal length, respectively, so their perimeters are 3, 4, 5, 6, and 8.

The median of these perimeters is the middle perimeter, 5. The midrange of these perimeters is the mean of the greatest and the least perimeters:

$\left ( 3+8\right ) \div 2 = 5.5$

The midrange, (B), is greater.

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Question

The length of a side of a regular octagon is one and a half times the hypotenuse of the above right triangle. Give the perimeter of the octagon in feet.

Answer

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.

The sidelength of the octagon is therefore

$50 \times 1\frac{1}{2} = 75$ inches,

and the perimeter of the regular octagon, which has eight sides of equal length, is

$75 \times 8 = 600$ inches,

or

$600 \div 12 = 50$ feet.

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Question

$\overline{PE}$ is a side of regular Pentagon as well as Square , which is completely outside Pentagon . $\overline{XY}$ is a side of equilateral $\bigtriangleup XYZ$ , where is a point outside Square . Which is the greater quantity?

(a) The perimeter of Pentagon

(b) The perimeter of Pentagon

Answer

The figure referenced is below:

Pentagon is regular, so all of its sides have the same length; we will examine $\overline{PE }$ in particular. The perimeter of Pentagon is the sum of the lengths of its sides, which is $5 \cdot PE$ .

Since $\overline{PE }$ is also a side of Square , it follows that ; since $\overline{XY }$ is also a side of equilateral $\bigtriangleup XYZ$ , . The perimeter of Pentagon is equal to

$= 5 \cdot PE$ ,

the same as that of Pentagon .

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Question

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

Which quantity is greater?

(a)

(b)

Answer

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

However, we cannot tell whether or is greater. For example, if , then ; if , then .

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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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