Triangles - ISEE Upper Level Mathematics Achievement

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Question

Which of the following is true about a triangle with two angles that measure $45^{\circ}$ and $45^{\circ}$ ?

Answer

The sum of the two given angles is 90 degrees, which means that the third angle should be a right angle (90 degrees). We also know that two of the angles are equal. Therefore, the triangle is right and isosceles.

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Question

What is the value of x in a right triangle if the two acute angles are equal to 5x and 25x?

Answer

In a right triangle, there is one right angle of 90 degrees, while the two acute angles add up to 90 degrees.

Given that the two acute angles are equal to 5x and 25x, the value of x can be calculated with the equation below:

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Question

One of the base angles of an isosceles triangle is $42^{\circ}$ . Give the measure of the vertex angle.

Answer

The base angles of an isosceles triangle are always equal. Therefore both base angles are $42^{\circ}$ .

Let the measure of the third angle. Since the sum of the angles of a triangle is $180^{\circ}$ , we can solve accordingly:

$42+42+x=180\Rightarrow x=180-84\Rightarrow x=96^{\circ}$

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Question

Which of the following is true about a triangle with two angles that measure $120^{\circ }$ and $90^{\circ }$ ?

Answer

A triangle must have at least two acute angles; however, a triangle with angles that measure $120^{\circ }$ and $90^{\circ }$ could have at most one acute angle, an impossible situation. Therefore, this triangle is nonexistent.

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Question

Which of the following is true about a triangle with two angles that measure $91^{\circ }$ each?

Answer

A triangle must have at least two acute angles; however, a triangle with angles that measure $91 ^{\circ }$ would have two obtuse angles and at most one acute angle. This is not possible, so this triangle cannot exist.

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Question

One angle of an isosceles triangle has measure $110^{\circ }$ . What are the measures of the other two angles?

Answer

An isosceles triangle not only has two sides of equal measure, it has two angles of equal measure. This means one of two things, which we examine separately:

Case 1: It has another $110^{\circ }$ angle. This is impossible, since a triangle cannot have two obtuse angles.

Case 2: Its other two angles are the ones that are of equal measure. If we let be their common measure, then, since the sum of the measures of a triangle is $180^{\circ }$ ,

$2x \div 2= 70 \div 2$

Both angles measure $35^{\circ}$

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Question

Note: Figure NOT drawn to scale.

What is the measure of angle

Answer

The two angles at bottom are marked as congruent. One forms a linear pair with a $152 ^{\circ }$ angle, so it is supplementary to that angle, making its measure $(180-152)^{\circ } = 28^{\circ }$ . Therefore, each marked angle measures $28^{\circ }$ .

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

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Question

The angles of a triangle measure $x^{\circ }, x^{\circ },\left ( x -42 \right ) ^{\circ }$ . Evaluate .

Answer

The sum of the degree measures of the angles of a triangle is 180, so we solve for in the following equation:

$3x \div 3 = 222 \div 3$

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Question

The acute angles of a right triangle measure $\left ( 2x - 8 \right )^{\circ }$ and $\left ( 3x - 4 \right )^{\circ }$ .

Evaluate .

Answer

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

$\left ( 2x - 8 \right ) + \left ( 3x - 4 \right ) = 90$

$5x \div 5 = 102 \div 5$

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Question

Note: Figure NOT drawn to scale

Refer to the above figure. $\overline{AB } \cong \overline{AC}$ ; $m \widehat{AC} = 140 ^{\circ }$ .

What is the measure of $\angle A$ ?

Answer

Congruent chords of a circle have congruent minor arcs, so since $\overline{AB } \cong \overline{AC}$ , $\widehat{AB } \cong \widehat{AC}$ , and their common measure is $m\widehat{AB } =m \widehat{AC} = 140^{\circ}$ .

Since there are $360^{\circ}$ in a circle,

$m\widehat{BC } + m\widehat{AB } + m \widehat{AC} = 360^{\circ}$

$m\widehat{BC } + 140^{\circ} + 140^{\circ}= 360^{\circ}$

$m\widehat{BC } + 280^{\circ}= 360^{\circ}$

$m\widehat{BC } = 80^{\circ}$

The inscribed angle $\angle A$ intercepts this arc and therefore has one-half its degree measure, which is $40 ^{\circ }$

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Question

Solve for :

Answer

The sum of the internal angles of a triangle is equal to . Therefore:

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Question

Figure NOT drawn to scale.

Refer to the above figure. Evaluate $m \angle ACB$ .

Answer

The measure of an exterior angle of a triangle, which here is $\angle ACD$ , is equal to the sum of the measures of its remote interior angles, which here are $\angle A$ and $\angle B$ . Consequently,

$m \angle A + m \angle B = m \angle ACD$

$6 \cdot 2 = 0.5x \cdot 2$

$m \angle ACD =( 6.5 x) ^{\circ }= (6.5 \cdot 12 )^{\circ }= 78^{\circ }$

$\angle ACB$ and $\angle ACD$ form a linear pair and, therefore,

$m \angle ACB=180 ^{\circ } - m \angle ACD = 180 ^{\circ } -78^{\circ } = 102 ^{\circ }$ .

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Question

Refer to the above figure. Express in terms of .

Answer

The measure of an interior angle of a triangle is equal to 180 degrees minus that of its adjacent exterior angle, so

$m \angle ABC = (180 - 82)^{\circ } = 98^{\circ }$

and

$m \angle ACB = (180 - x)^{\circ }$ .

The sum of the degree measures of the three interior angles is 180, so

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

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Question

In the above figure, $\overline{AC} \cong \overline{AD}$ .

Give the measure of $\angle CAD$ .

Answer

$\angle BCA$ and $\angle ACD$ form a linear pair, so their degree measures total $180 ^{\circ }$ ; consequently,

$m \angle ACD = 180 ^{\circ } - m \angle CAD$

$= 180 ^{\circ } - 108 ^{\circ }$

$= 72 ^{\circ }$

$\overline{AC} \cong \overline{AD}$ , so by the Isosceles Triangle Theorem,

$m \angle ADC = m \angle ACD = 72 ^{\circ }$

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

$\angle CAD + m \angle ADC + m \angle ACD = 180 ^{\circ }$

$\angle CAD + 72^{\circ } +72^{\circ } = 180 ^{\circ }$

$\angle CAD +144^{\circ } = 180 ^{\circ }$

$\angle CAD = 36 ^{\circ }$

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Question

Which of the following is true about a triangle with two angles that measure $120^{\circ }$ and $90^{\circ }$ ?

Answer

A triangle must have at least two acute angles; however, a triangle with angles that measure $120^{\circ }$ and $90^{\circ }$ could have at most one acute angle, an impossible situation. Therefore, this triangle is nonexistent.

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Question

$\Delta ABC \sim \Delta DEF$

$\angle B$ is a right angle; , .

Which is the greater quantity?

(a) $m \angle A$

(b) $60 ^{\circ }$

Answer

$\Delta ABC \sim \Delta DEF$ . Corresponding angles of similar triangles are congruent, so since $\angle B$ is a right angle, so is $\angle E$ .

The hypotenuse $\overline{DF}$ of $\Delta DEF$ is twice as long as leg $\overline{DE}$ ; by the $30 ^{\circ } - 60^{\circ } -90^{\circ }$ Theorem, $m \angle D = 60 ^{\circ }$ . Again, by similiarity,

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

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Question

NOTE: Figures NOT drawn to scale.

Refer to the above two triangles.

$\Delta ABC \sim \Delta DEF ,AB = 25, DE = 20, AC = 20$

What is ?

Answer

Corresponding sides of similar triangle are proportional, so if

$\Delta ABC \sim \Delta DEF$ , then

$\frac{DF}{AC} = \frac{DE}{AB}$

Substitute the known sidelengths, then solve for :

$\frac{DF}{20} = \frac{20}{25}$

$\frac{DF}{20}\cdot 20= \frac{20}{25}\cdot 20$

$DF = \frac{400}{25}$

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Question

$\Delta ABC \sim \Delta DEF$

What is the perimeter of $\Delta DEF$ ?

Answer

By definition, since, $\Delta ABC \sim \Delta DEF$ , side lengths are in proportion.

So,

$\frac{DE}{AB}= \frac{DF}{AC}$

$\frac{DE}{15}= \frac{36}{12} = 3$

$\frac{DE}{15} \cdot 15 = 3\cdot 15$

$\frac{EF}{BC}= \frac{DF}{AC}$

$\frac{EF}{21}= \frac{36}{12} = 3$

$\frac{EF}{21} \cdot 21 = 3\cdot 21$

The perimeter of $\Delta DEF$ is

.

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Question

$\Delta ABC \sim \Delta DEF$

What is ?

Answer

By definition, since $\Delta ABC \sim \Delta DEF$ , all side lengths are in proportion.

$\frac{AC}{DF} = \frac{AB}{DE}$

$\frac{AC}{66} = \frac{15}{45}$

$\frac{AC}{66} = \frac{1}{3}$

$AC = \frac{1}{3} \cdot 66 = 22$

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Question

One angle of a right triangle measures 45 degrees, and the hypotenuse measures 8 centimeters. Give the area of the triangle.

Answer

This triangle has two angles of 45 and 90 degrees, so the third angle must measure 45 degrees; this is therefore an isosceles right triangle.

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 8^2=2s^2\Rightarrow s^2=32$

$\Rightarrow s=\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\ cm$

We can then plug this side length into the formula for area.

$area=\frac{s^2}{2}=\frac{(4\sqrt{2})^2}{2}=\frac{32}{2}=16\ cm^2$

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