Acute / Obtuse 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 $120^{\circ }$ 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

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

Two sides of a scalene triangle measure 4 centimeters and 7 centimeters, and their corresponding angle measures 30 degrees. Find the area of the triangle.

Answer

$Area=\frac{1}{2}absinC$ ,

where and are the lengths of two sides and is the angle measure.

Plug in our given values:

$Area=\frac{1}{2}absinC=\frac{1}{2}\times 4\times 7\times sin30^{\circ}$

$sin30^{\circ}=\frac{1}{2}\Rightarrow Area=\frac{1}{2}\times 4\times 7\times \frac{1}{2}=7\ cm^2$

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Question

A scalene triangle has a base length and a corresponding altitude of . Give the area of the triangle in terms of .

Answer

$Area =\frac{bh}2{}$ ,

where is the base and is the altitude.

$Area =\frac{bh}2{}=\frac{t\times 4t}{2}=2t^2$

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Question

What is the area of a triangle on the coordinate plane with its vertices on the points ?

Answer

The base can be seen as the (horizontal) line segment connecting $\left (-5,0 \right )$ and $\left (10,0 \right )$ , the length of which is . The height is the pependicular distance from to the segment; since the segment is part of the -axis, this altitude is vertical and has a length equal to -coordinate .

The area of this triangle is therefore

$A = \frac{1}{2} bh = \frac{1}{2} \cdot15 \cdot 8 = 60$ .

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Question

What is the area of a triangle on the coordinate plane with its vertices on the points ?

Answer

The base can be seen as the (vertical) line segment connecting and , which has length . The height is the pependicular distance from to the segment; since the segment is part of the -axis, this altitude is horizontal and has length equal to -coordinate .

The area of this triangle is therefore

$A = \frac{1}{2} bh = \frac{1}{2} \cdot13 \cdot 6 = 39$ .

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Question

Figure NOT drawn to scale.

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . What percent of $\bigtriangleup ABC$ has been shaded gray?

Choose the closest answer.

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other as well as the large triangle.

The similarity ratio of $\bigtriangleup AXB$ to $\bigtriangleup ABC$ is the ratio of the lengths of their hypotenuses. The hypotenuse of the latter is 18; that of the former, from the Pythagorean Theorem, is

$\sqrt{18 ^{2}+ 30 ^{2}} = \sqrt{324+ 900} = \sqrt{1,224}$

The similarity ratio is therefore $\frac{18}{ \sqrt{1,224}}$ . The ratio of their areas is the square of this, or

$\left (\frac{18}{ \sqrt{1,224}} \right )^{2} = \frac{324}{1,224} = \frac{324 \div 36 }{1,224\div 36} = \frac{9}{34}$

The area of $\bigtriangleup AXB$ is

$\frac{9}{34} \times 100 % \approx 26.5 %$

of that of $\bigtriangleup ABC$ , so the choice closest to the correct percent is 25%.

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