Acute / Obtuse Triangles

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ISEE Upper Level Quantitative Reasoning › Acute / Obtuse Triangles

Questions 1 - 10

Triangle

Figure NOT drawn to scale.

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

$102 ^{\circ }$

$97^{\circ }$

$107^{\circ }$

$112^{\circ }$

Explanation

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,

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

Triangle

Figure NOT drawn to scale.

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

$102 ^{\circ }$

$97^{\circ }$

$107^{\circ }$

$112^{\circ }$

Explanation

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

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

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

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

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

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

Evaluate .

Explanation

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

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

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

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

Evaluate .

Explanation

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

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

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

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.

$7\ cm^2$

$10\ cm^2$

$8\ cm^2$

$5\ cm^2$

Explanation

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

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

Plug in our given values:

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

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

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

Explanation

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

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

$\Delta ABC \sim \Delta DEF$

What is ?

It is impossible to tell from the information given.

Explanation

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

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

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

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

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

Right triangle

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.

Explanation

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%.

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.

$7\ cm^2$

$10\ cm^2$

$8\ cm^2$

$5\ cm^2$

Explanation

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

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

Plug in our given values:

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

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

Right triangle

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.

Explanation

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.

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

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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