Acute / Obtuse Triangles - SSAT Upper Level Quantitative

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If the vertex angle of an isoceles triangle is , what is the value of one of its base angles?

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In an isosceles triangle, the base angles are the same. Also, the three angles of a triangle add up to .

So, subtract the vertex angle from . You get .

Because there are two base angles you divide by , and you get .

Figure NOT drawn to scale.

If $x = $72^{\circ }$$ and $y = $43^{\circ }$$ , evaluate .

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The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so

$w = x + y = 72 + 43 = 115 ^{\circ }$

Note: Figure NOT drawn to scale.

Refer to the above diagram.

Which of the following could be a measure of $\angle 3$ ?

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The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so

$\angle 3 = \angle 1 + \angle 2$ .

We also have the following constraints:

Then, by the addition property of inequalities,

$105 ^{\circ }\leq m \angle 3 \leq $120^{\circ }$$

Therefore, the measure of $\angle 3$ must fall in that range. Of the given choices, only falls in that range.

Refer to the above diagram.

Which of the following could be a measure of $\angle 2$ ?

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The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so

$\angle 3 = \angle 1 + \angle 2$

or

$\angle 2 = \angle 3 - \angle 1$

Therefore, the maximum value of $\angle 2$ is the least possible value of $\angle 1$ subtracted from the greatest possible value of $\angle 3$ :

The minimum value of $\angle 2$ is the greatest possible value of $\angle 1$ subtracted from the least possible value of $\angle 3$ :

Therefore,

Since all of the choices fall in this range, all are possible measures of $\angle 2$ .

Find the angle measurement of .

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All the angles in a triangle must add up to .

Find the angle measurement of .

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All the angles in a triangle must add up to

Find the angle measurement of .

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All the angles in a triangle must add up to .

The interior angles of a triangle measure . Of these three degree measures, give the greatest.

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The degree measures of the interior angles of a triangle total 180 degrees, so

$\frac{4x}{4} = \frac{153}{4}$

$x = 38 \frac{1}{4}$

One angle measures $38\frac{1}{4}^{\circ }$

The other two angles measure

$x+ 26 = 38\frac{1}{4} + 26 = 64\frac{1}{4} ^{\circ }$

and

$2x+ 1 = 2 \left\(38\frac{1}{4}\right\)+ 1 = 77 \frac{1}{2} ^{\circ }$ .

We want the greatest of the three, or $77 \frac{1}{2} ^{\circ }$ .

An isosceles triangle has an angle whose measure is .

What could be the measures of one of its other angles?

(a)

(b)

(c)

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By the Isosceles Triangle Theorem, an isosceles triangle has two congruent interior angles. There are two possible scenarios if one angle has measure :

Scenario 1: The other two angles are congruent to each other. The degree measures of the interior angles of a triangle total , so if we let be the common measure of those angles:

This makes (b) a possible answer.

Scenario 2: One of the other angles measures also, making (c) a possible answer. The degree measure of the third angle is

,

making (a) a possible answer. Therefore, the correct choice is (a), (b), or (c).

One of the interior angles of a scalene triangle measures . Which of the following could be the measure of another of its interior angles?

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A scalene triangle has three sides of different measure, so, by way of the Converse of the Isosceles Triangle Theorem, each angle is of different measure as well. We can therefore eliminate immediately.

Also, if the triangle also has a angle, then, since the total of the degree measures of the angles is , it follows that the third angle has measure

$180 - (54+72) = 180 - 126 = 54 ^{\circ }$ .

Therefore, the triangle has two angles that measure the same, and can be eliminated.

Similarly, if the triangle also has a angle, then, since the total of the degree measures of the angles is , it follows that the third angle has measure

$180 - (54+63) = 180 - 117= 63 ^{\circ }$ .

The triangle has two angles that measure . This choice can be eliminated.

can be eliminated, since the third angle would have measure

$180 - (54+126) = 180 - 180= $0^{\circ }$$ ,

an impossible situation since angle measures must be positive.

The remaining possibility is . This would mean that the third angle has measure

$180 - (54+108) = 180 - 162= $18^{\circ }$$ .

The three angles have different measures, so the triangle is scalene. is the correct choice.

Given: $\bigtriangleup ABC$ with $m \angle A = $20^{\circ }$, m \angle B = $32^{\circ }$$ . Locate on $\overline{AB}$ so that $\overrightarrow{CD}$ is the angle bisector of $\angle ACB$ . What is $m \angle CDB$ ?

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Above is the figure described.

The measures of the interior angles of a triangle total , so the measure of $\angle ACB$ is

$m \angle ACB = $180^{\circ }$ -( m \angle A + m \angle B)$

$= $180^{\circ }$ -( $20^{\circ }$ + $32^{\circ }$)$

$= $180^{\circ }$ - $52^{\circ }$$

$= $128^{\circ }$$

Since $\overrightarrow{CD}$ bisects this angle,

$m \angle BCD= $\frac{1}{2}$m \angle ACB = $\frac{1}{2}$ \cdot $128^{\circ }$ $=64^{\circ }$$

and

$m \angle CDB = $180^{\circ }$ -( m \angle BCD + m \angle B)$

$= $180^{\circ }$ -( 64 ^{\circ } + $32^{\circ }$ )$

$= $180^{\circ }$ $-96^{\circ }$$

$= $84^{\circ }$$

Given: $\bigtriangleup ABC$ with $m \angle B = $74^{\circ }$$ . is located on $\overline{AB}$ so that $\overrightarrow{CD}$ bisects $\angleACB$ $\angle ACB$ and forms isosceles triangle $\bigtriangleup BCD$ .

Give the measure of $\angle A$ .

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If $\bigtriangleup BCD$ is isosceles, then by the Isosceles Triangle Theorem, two of its angles must be congruent.

Case 1: $m \angle B = m \angle BCD= $74^{\circ }$$

Since $\overrightarrow{CD}$ bisects $\angleACB$ $\angle ACB$ into two congruent angles, one of which must be $m \angle BCD$ ,

$m \angle ACB = 2 \cdot m \angle BCD = 2 \cdot $74^{\circ }$ = $148^{\circ }$$

However, this is impossible, since $\angle ACB$ and $\angle B$ are two angles of the original triangle; their total measure is

$m \angle ACB+ m \angle B = $148^{\circ }$+ 74 ^{\circ } = 222 ^{\circ } > 180 ^{\circ }$

Case 2: $m \angle B = m \angle BDC= $74^{\circ }$$

Then, since the degree measures of the interior angles of a triangle total ,

$m \angle BCD= $180^{\circ }$ - ( \angle B + m \angle BDC )$

$= $180^{\circ }$ - $(74^{\circ }$ + $74^{\circ }$ )$

$= 180 ^{\circ } - $148^{\circ }$$

$= $32^{\circ }$$

Since $\overrightarrow{CD}$ bisects $\angleACB$ $\angle ACB$ into two congruent angles, one of which must be $m \angle BCD$ ,

$m \angle ACB = 2 \cdot m \angle BCD = 2 \cdot $32^{\circ }$ = $64^{\circ }$$

and

$m \angle A = $180^{\circ }$ - ( \angle B + m \angle ACB )$

$= $180^{\circ }$ - $(74^{\circ }$ + $64^{\circ }$ )$

$= $180^{\circ }$ - $138^{\circ }$$

$= $42^{\circ }$$

Case 3: $m \angle BDC= m \angle BCD$

Then

$m \angle B = $180^{\circ }$ -(m \angle BDC+ m \angle BCD)$

$74= $180^{\circ }$ -(m \angle BCD+ m \angle BCD)$

$74= $180^{\circ }$ -2 \cdot m \angle BCD$

$2 \cdot m \angle BCD = $106^{\circ }$$

$m \angle BCD = $53^{\circ }$$

$m \angle ACB = 2 \cdot m \angle BCD = 2 \cdot $53^{\circ }$ = $106^{\circ }$$

$m \angle A = $180^{\circ }$ - ( \angle B + m \angle ACB )$

$= $180^{\circ }$ - ( $74^{\circ }$+ $106^{\circ }$ )$

$= $0^{\circ }$$ , which is not possible.

Therefore, the only possible measure of $\angle A$ is .

$\bigtriangleup ABC$ is a right triangle with right angle $\angle A$ . is located on $\overline{AB}$ so that, when $\overline{CD}$ is constructed, isosceles triangles $\bigtriangleup ADC$ and $\bigtriangleup BDC$ are formed.

What is the measure of $\angle B$ ?

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The figure referenced is below:

Since $\bigtriangleup ADC$ is an isosceles right triangle, its acute angles - in particular, $\angle ADC$ - measure each. Since this angle forms a linear pair with $\angle CDB$ :

$m \angle CDB = $180^{\circ }$ - m \angle ADC = $180^{\circ }$ - 45 ^{\circ } = $135^{\circ }$$ .

$\bigtriangleup BDC$ is also isosceles, so, by the Isosceles Triangle Theorem, it has two congruent angles. Since $\angle CDB$ is obtuse, and no triangle has two obtuse angles:

$m \angle BCD = m \angle B$ .

Also, $\angle ADC$ is an exterior angle of $\bigtriangleup BDC$ , whose measure is equal to the sum of those of its two remote interior angles, which are the congruent angles $m \angle BCD = m \angle B$ . Therefore,

$m \angle ADC = m \angle BCD + m \angle B$

$m \angle B = $22$\frac{1}{2}$^{\circ }$$

If the vertex angle of an isosceles triangle is 30 degrees, what is the value of the angle that is exterior to one of the base angles?

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First, you subtract one angle so . Because there are 2 base angles you divide that value by 2 to get 75 Degrees. The exterior angle and the base angle are supplementary angles so .

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; $\bigtriangleup DEF$ has perimeter 400.

Which of the following is equal to $DF \div DE$ ?

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The perimeter of $\bigtriangleup DEF$ is actually irrelevant to this problem. Corresponding sides of similar triangles are in proportion, so use this to calculate $DF \div DE$ , or $\frac{DF}{DE}$ :

$$\frac{DF}{DE}$ = $\frac{AC}{AB}$ = $\frac{36}{24}$ = 1.5$

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; $\bigtriangleup DEF$ has perimeter 300.

Evaluate .

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The ratio of the perimeters of two similar triangles is equal to the ratio of the lengths of a pair of corresponding sides. Therefore,

$$\frac{DE}{AB}$= $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$ and $$\frac{EF}{BC}$= $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$ , or

$$\frac{DE}{AB}$=\frac{EF}{BC}$ = $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$

By one of the properties of proportions, it follows that

$$\frac{DE+EF}{AB+BC}$ = $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$

The perimeter of $\bigtriangleup ABC$ is

, so

$\frac{DE+EF}{24+20}$ = $\frac{300}{80}$

$\frac{DE+EF}{44}$ = $\frac{300}{80}$

$$\frac{DE+EF}{44}$ \cdot 44 = $\frac{300}{80}$ \cdot 44$

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; ; $\bigtriangleup DEF$ has perimeter 90.

Give the perimeter of $\bigtriangleup ABC$ .

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The ratio of the perimeters of two similar triangles is the same as the ratio of the lengths of a pair of corresponding sides. Therefore,

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

$m\angle A + m\angle D = $94^{\circ }$$

$m\angle B + m\angle F = $94^{\circ }$$

Evaluate $m \angle C + m \angle E$ .

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The similarity of the triangles is actually extraneous information here. The sum of the measures of a triangle is $180 ^{\circ }$ , so:

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

$m \angle D + m \angle E + m \angle F = $180^{\circ }$$

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

$m \angle C + m \angle E + (m \angle A +m \angle D) + (m \angle B + m \angle F )= $360^{\circ }$$

$m \angle C + m \angle E +94 ^{\circ } +94 ^{\circ }= $360^{\circ }$$

$m \angle C + m \angle E +188 ^{\circ }= $360^{\circ }$$

$m \angle C + m \angle E +188 ^{\circ }- $188^{\circ }$ = $360^{\circ }$ - $188^{\circ }$$

$m \angle C + m \angle E = $172^{\circ }$$

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ ; $\frac{AB}{DE}$ = $\frac{AC}{DF}$ .

Which of the following statements would not be enough, along with what is given, to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

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From both the given proportion statement and either $\frac{AB}{DE}$ = $\frac{BC}{EF}$ or $\frac{AC}{DF}$= $\frac{BC}{EF}$ , it follows that $\frac{AB}{DE}$ =\frac{AC}{DF}$= $\frac{BC}{EF}$ —all three pairs of corresponding sides are in proportion; by the Side-Side-Side Similarity Theorem, $\bigtriangleup ABC \sim \bigtriangleup DEF$ . From the given proportion statement and $\angle A \cong \angle D$ , since these are the included angles of the sides that are in proportion, then by the Side-Angle-Side Similarity Theorem, $\bigtriangleup ABC \sim \bigtriangleup DEF$ . From the given proportion statement and $\angle C \cong \angle F$ , since these are nonincluded angles of the sides that are in proportion, no similarity can be deduced.

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ ; $\angle A \cong \angle D$ and $\angle C \cong \angle F$ .

Which of the following statements would not be enough, along with what is given, to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

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Two pairs of corresponding angles are stated to be congruent in the main body of the problem; it follows from the Angle-Angle Similarity Postulate that the triangles are similar. No further information is needed.