Plane Geometry - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

Which is the greater quantity?

(a) The length of the line segment connecting and

(b) The length of the line segment connecting and

Answer

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

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

(b) The length of the line segment connecting and 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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Question

In isosceles triangle ABC, the measure of angle A is 50 degrees. Which is NOT a possible measure for angle B?

Answer

If angle A is one of the base angles, then the other base angle must measure 50 degrees. Since 50 + 50 + x = 180 means x = 80, the vertex angle must measure 80 degrees.

If angle A is the vertex angle, the two base angles must be equal. Since 50 + x + x = 180 means x = 65, the two base angles must measure 65 degrees.

The only number given that is not possible is 95 degrees.

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Question

Let the three angles of a triangle measure , , and .

Which of the following expressions is equal to ?

Answer

The sum of the measures of the angles of a triangle is $180^{\circ }$ , so simplify and solve for in the equation:

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Question

The angles of a triangle measure $\left ( x +30 \right )^{\circ }$ , $\left ( 2x - 3\right ) ^{\circ }$ , and $y ^{\circ }$ . Give in terms of .

Answer

The sum of the measures of three angles of a triangle is $180 ^{\circ }$ , so we can set up the equation:

$\left ( x +30 \right )+\left ( 2x - 3\right ) + y = 180$

We can simplify and solve for :

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Question

Which of the following is true about a triangle with two angles that measure $44^{\circ }$ each?

Answer

The measures of the angles of a triangle total $180^{\circ }$ , so if two angles measure $44^{\circ }$ and we call the measure of the third, then

$x= 92^{\circ } > 90^{\circ }$

This makes the triangle obtuse.

Also, since the triangle has two congruent angles (the $44^{\circ }$ angles), the triangle is also isosceles.

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Question

You are given two triangles, $\Delta ABC$ and $\Delta DEF$ .

, $\angle B$ is an acute angle, and $\angle E$ is a right angle.

Which quantity is greater?

(a)

(b)

Answer

We invoke the SAS Inequality Theorem, which states that, given two triangles $\Delta ABC$ and $\Delta DEF$ , with , $m\angle B < m \angle E$ ( the included angles), then - that is, the side opposite the greater angle has the greater length. Since $\angle B$ is an acute angle, and $\angle E$ is a right angle, we have just this situation. This makes (b) the greater.

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Question

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

The Triangle Exterior-Angle Theorem states that the measure of an exterior angle is equal to the sum of its remote interior angles. Therefore,

,

making the quantities equal.

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

(a) The measures of the angles of a linear pair total 180, so:

(b) The Triangle Exterior-Angle Theorem states that the measure of an exterior angle is equal to the sum of its remote interior angles. Therefore, .

Therefore (a) is the greater quantity.

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

The two angles at bottom are marked as congruent. Each of these two angles forms a linear pair with a $140 ^{\circ }$ angle, so it is supplementary to that angle, making its measure $(180-140)^{\circ } = 40^{\circ }$ . Therefore, the other marked angle also measures $40^{\circ }$ .

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so:

The quantities are equal.

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Question

$\Delta ABC$ is equilateral; $\Delta CBD$ is isosceles

Which is the greater quantity?

(a) $m \angle BCD$

(b) $60 ^{\circ }$

Answer

$\Delta ABC$ is equilateral, so

.

In $\Delta CBD$ , we are given that

.

Since the triangles have two pair of congruent sides, the third side with the greater length is opposite the angle of greater measure. Therefore,

$m \angle BCD >m \angle BAC$ .

Since $\angle BAC$ is an angle of an equilateral triangle, its measure is $60^{\circ }$ , so $m \angle BCD >60^{\circ }$ .

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$

$m \angle A = 60^{\circ }$

$\angle E \cong \angle F$

Which is the greater quantity?

(a) $m \angle C$

(b) $m \angle D$

Answer

Corresponding angles of similar triangles are congruent, so, since $\bigtriangleup ABC \sim \bigtriangleup DEF$ , it follows that

$m \angle D = m \angle A = 60 ^{\circ }$

By similarity, $m \angle B = m \angle E$ and $m \angle C = m \angle F$ , and we are given that $m \angle E = m \angle F$ , so

$m \angle B = m \angle C$

Also,

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

$60^{\circ }+ m \angle C + m \angle C = 180^{\circ }$

$60^{\circ }+2 \cdot m \angle C = 180^{\circ }$

$2 \cdot m \angle C = 120^{\circ }$

$m \angle C = 60^{\circ }$ ,

and $m \angle C =m \angle D$ .

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Question

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

Extend $\overline{AB}$ as seen in the figure below:

The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles; specifically,

$m \angle CAD = m \angle B + m \angle C$ ,

and

$m \angle CAD =( x+ y) ^{\circ }$

However, $m \angle CAD > z ^{\circ }$ , so, by substitution,

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Question

Given: $\bigtriangleup ABC$ . . Which is the greater quantity?

(a) $m \angle B$

(b) $60^{\circ }$

Answer

Below is the referenced triangle along with $\bigtriangleup DEF$ , an equilateral triangle with sides of length 10:

As an angle of an equilateral triangle, $\angle D$ has measure $60 ^{\circ }$ . Applying the Side-Side-Side Inequality Theorem, since , , and , it follows that $m \angle A > m \angle D$ , so $m \angle A > 60^{\circ }$ .

Also, since , by the Isosceles Triangle Theorem, $m \angle B = m \angle C$ . Since $m \angle A > 60^{\circ }$ , and the sum of the measures of the angles of a triangle is $180 ^{\circ }$ , it follows that

$m \angle B + m \angle C < 120^{\circ }$

Substituting and solving:

$m \angle B + m \angle B < 120^{\circ }$

$2 m \angle B < 120^{\circ }$

$2 m \angle B \div 2 < 120^{\circ } \div 2$

$m \angle B < 60^{\circ }$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure.

Which is the greater quantity?

(a)

(b) $30 ^{\circ }$

Answer

Since the shorter leg of the right triangle is half the hypotenuse, the triangle is a $30 ^{\circ } -60^{\circ } -90^{\circ }$ triangle, with the $30 ^{\circ }$ angle opposite the shorter leg. That makes .

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Question

Right triangle $\Delta ABC$ has right angle $\angle B$ .

$m\angle A = \left ( x + 30 \right )^ {\circ} ,m\angle C = \left ( 2x- 57\right ) ^ {\circ}$

Which is the greater quantity?

(a) $m\angle A$

(b) $m\angle C$

Answer

The degree measures of the acute angles of a right triangle total 90, so we solve for in the following equation:

$\left ( x + 30 \right )+ \left ( 2x- 57\right ) = 90$

$3x \div 3 = 117 \div 3$

(a) $m\angle A = \left ( x + 30 \right )^ {\circ} = \left ( 39 + 30 \right )^ {\circ} = 69^ {\circ}$

(b) $m\angle C =\left ( 90 - 69 \right ) ^ {\circ} = 21^ {\circ}$

$m\angle A > m\angle C$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$

$\angle A$ is a right angle.

Which is the greater quantity?

(a) $m \angle D$

(b) $m \angle E$

Answer

Corresponding angles of similar triangles are congruent, so, since $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and $\angle A$ is right, it follows that

$m \angle D = m \angle A = 90 ^{\circ }$

$\angle D$ is a right angle of a right triangle $\bigtriangleup DEF$ . The other two angles must be acute - that is, with measure less than $90 ^{\circ }$ - so $m \angle D > m \angle E$ .

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Question

$\bigtriangleup ABC$ is inscribed in a circle. $\angle B$ is a right angle, and $m \overarc{ACB} = 270 ^{\circ }$ .

Which is the greater quantity?

(a) $m \angle A$

(b) $m \angle C$

Answer

The figure referenced is below:

$\overarc{ACB}$ has measure $270 ^{\circ }$ , so its corresponding minor arc, $\overarc{A B}$ , has measure $360 ^{\circ } - 270 ^{\circ } = 90^{\circ }$ . The inscribed angle that intercepts this arc, which is $\angle C$ , has measure half this, or $45 ^{\circ }$ . Since $\angle B$ is a right angle, the other acute angle, $\angle A$ , has measure

$m \angle A = 90 ^{\circ } - m \angle C = 90 ^{\circ } - 45^{\circ } = 45 ^{\circ }$

Therefore, $m \angle A = 45 ^{\circ } = m \angle C$ .

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Question

Consider a triangle, $\bigtriangleup ABC$ , in which , , and . Which is the greater number?

(a) The measure of $\angle B$ in degrees

(b)

Answer

By the Converse of the Pythagorean Theorem, a triangle is right if and only if the sum of the squares of the lengths of the smallest two sides is equal to the square of the longest side. Compare the quantities $(AB)^{2}+ ( BC )^{2}$ and $(AC)^{2}$

$(AB)^{2}+ ( BC )^{2} = 36 ^{2} + 48^{2} = 1,296 + 2,304 = 3,600$

$(AC)^{2} = 60^{2} = 3,600$

$(AB)^{2}+ ( BC )^{2} =(AC)^{2}$ , so $\bigtriangleup ABC$ is right, with the right angle opposite longest side $\overline{AC}$ . Thus, $\angle B$ is right and has degree measure 90.

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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

Which is the greater quantity?

(a) The measure of an angle complementary to a $55 ^{\circ }$ angle

(b) The measure of an angle supplementary to a $135 ^{\circ }$ angle

Answer

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

(a) The measure of an angle complementary to a $55 ^{\circ }$ angle is $90^{\circ } - 55^{\circ } = 35^{\circ }$

(b) The measure of an angle supplementary to a $135 ^{\circ }$ angle is $180^{\circ } - 135 ^{\circ } = 45^{\circ }$

This makes (b) greater.

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