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Intermediate Geometry : Plane Geometry

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Example Questions

Example Question #131 : Acute / Obtuse Triangles

Given: $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ and $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$.

AB = 10 $\displaystyle AB = 10$

BC = 15 $\displaystyle BC = 15$

DE = 12 $\displaystyle DE = 12$

EF = 18 $\displaystyle EF = 18$

$m\angle B = 78 ^{\circ }$ $\displaystyle m\angle B = 78 ^{\circ }$

$m\angle E = 78 ^{\circ }$ $\displaystyle m\angle E = 78 ^{\circ }$

True or false: It follows from the given information that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

Possible Answers:

False

True

Correct answer:

True

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$, then $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ correspond respectively to $\displaystyle D$, $\displaystyle E$, and $\displaystyle F$.

According to the Side-Angle-Side Similarity Theorem (SASS), if the lengths of two pairs of corresponding sides of two triangles are in proportion, and their included angles are congruent, then the triangles are similar.

$\overline{AB}$ $\displaystyle \overline{AB}$ and $\overline{DE}$ $\displaystyle \overline{DE}$ are corresponding sides, as are $\overline{BC}$ $\displaystyle \overline{BC}$ and $\overline{EF}$ $\displaystyle \overline{EF}$; $\angle B$ $\displaystyle \angle B$ and $\angle E$ $\displaystyle \angle E$ are their included angles. Substituting

$\frac{AB}{DE} = \frac{10}{12}= \frac{5}{6}$ $\displaystyle \frac{AB}{DE} = \frac{10}{12}= \frac{5}{6}$

$\frac{BC}{EF} = \frac{15}{18} = \frac{5}{6}$ $\displaystyle \frac{BC}{EF} = \frac{15}{18} = \frac{5}{6}$

Therefore, $\frac{AB}{DE} =\frac{BC}{EF}$ $\displaystyle \frac{AB}{DE} =\frac{BC}{EF}$, and corresponding sides are in proportion.

$m \angle B = 78^{\circ }$ $\displaystyle m \angle B = 78^{\circ }$ and $m \angle E = 78^{\circ }$ $\displaystyle m \angle E = 78^{\circ }$; the included angles are congruent.

The conditions of SASS are met, and it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

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Example Question #132 : Acute / Obtuse Triangles

Given: $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ and $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$.

$\displaystyle A B = 12$

AC = 18 $\displaystyle AC = 18$

BC = 24 $\displaystyle BC = 24$

DE = 18 $\displaystyle DE = 18$

DF = 24 $\displaystyle DF = 24$

EF = 30 $\displaystyle EF = 30$

True or false: From the above six statements, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

Possible Answers:

True

False

Correct answer:

False

Explanation:

If $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$, then corresponding sides must be in proportion; that is, it must hold that

$\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$ $\displaystyle \frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$

Substituting the lengths of the sides for the respective quantities:

$\displaystyle A B = 12$

AC = 18 $\displaystyle AC = 18$

BC = 24 $\displaystyle BC = 24$

DE = 18 $\displaystyle DE = 18$

DF = 24 $\displaystyle DF = 24$

EF = 30 $\displaystyle EF = 30$

$\frac{AB}{DE} = \frac{12}{18} = \frac{2}{3}$ $\displaystyle \frac{AB}{DE} = \frac{12}{18} = \frac{2}{3}$

$\frac{AC}{DF} = \frac{18}{30} = \frac{3}{5}$ $\displaystyle \frac{AC}{DF} = \frac{18}{30} = \frac{3}{5}$

The inequality of these two side ratios disproves the similarity of the triangles, so the correct answer is "false".

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Example Question #133 : Acute / Obtuse Triangles

Given: $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ and $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$.

$m \angle A = 56 ^{\circ }$ $\displaystyle m \angle A = 56 ^{\circ }$

$m \angle B = 72 ^{\circ }$ $\displaystyle m \angle B = 72 ^{\circ }$

$m \angle E = 72^{\circ }$ $\displaystyle m \angle E = 72^{\circ }$

$m \angle F = 52^{\circ }$ $\displaystyle m \angle F = 52^{\circ }$

True or false: From the above four statements, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

Possible Answers:

False

True

Correct answer:

True

Explanation:

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ $\displaystyle 180^{\circ }$. Therefore,

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

Set $m \angle A = 56 ^{\circ }$ $\displaystyle m \angle A = 56 ^{\circ }$ and $m \angle B = 72 ^{\circ }$ $\displaystyle m \angle B = 72 ^{\circ }$, and solve for $m \angle C$ $\displaystyle m \angle C$:

$56 ^{\circ } + 72 ^{\circ } + m \angle C = 180^{\circ }$ $\displaystyle 56 ^{\circ } + 72 ^{\circ } + m \angle C = 180^{\circ }$

$128 ^{\circ } + m \angle C = 180^{\circ }$ $\displaystyle 128 ^{\circ } + m \angle C = 180^{\circ }$

$128 ^{\circ } + m \angle C - 128 ^{\circ } = 180^{\circ } - 128 ^{\circ }$ $\displaystyle 128 ^{\circ } + m \angle C - 128 ^{\circ } = 180^{\circ } - 128 ^{\circ }$

$m \angle C = 52 ^{\circ }$ $\displaystyle m \angle C = 52 ^{\circ }$

By the Angle-Angle Similarity Postulate (AA), two triangles are similar if two angles of the first triangle are congruent to those of their counterparts in the second. $m \angle B = 72 ^{\circ }$ $\displaystyle m \angle B = 72 ^{\circ }$ and $m \angle E = 72^{\circ }$ $\displaystyle m \angle E = 72^{\circ }$, so $\angle B \cong \angle E$ $\displaystyle \angle B \cong \angle E$; $m \angle C = 52 ^{\circ }$ $\displaystyle m \angle C = 52 ^{\circ }$ and $m \angle F = 52^{\circ }$ $\displaystyle m \angle F = 52^{\circ }$, so $\angle C \cong \angle F$ $\displaystyle \angle C \cong \angle F$. The conditions of AA are met, so it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$. The correct response is "true".

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Example Question #134 : Acute / Obtuse Triangles

Given: $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ and $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$.

$\angle A \cong \angle C$ $\displaystyle \angle A \cong \angle C$

$\angle B \cong \angle E$ $\displaystyle \angle B \cong \angle E$

$\angle D \cong \angle F$ $\displaystyle \angle D \cong \angle F$

True or false: From the above three statements, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

Possible Answers:

False

True

Correct answer:

True

Explanation:

Let $x = m \angle B = m \angle E$ $\displaystyle x = m \angle B = m \angle E$.

Let $y = m \angle A = m \angle C$ $\displaystyle y = m \angle A = m \angle C$

The measures of the interior angles of a triangle total $180^{\circ }$ $\displaystyle 180^{\circ }$. Therefore,

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

Substituting:

$y + x + y= 180^{\circ }$ $\displaystyle y + x + y= 180^{\circ }$

$2 y + x = 180^{\circ }$ $\displaystyle 2 y + x = 180^{\circ }$

$2 y + x - x = 180^{\circ }- x$ $\displaystyle 2 y + x - x = 180^{\circ }- x$

$2 y = 180^{\circ }- x$ $\displaystyle 2 y = 180^{\circ }- x$

$\frac{2y}{2} = \frac{180^{\circ }- x}{2}$ $\displaystyle \frac{2y}{2} = \frac{180^{\circ }- x}{2}$

$y = \frac{180^{\circ }- x}{2}$ $\displaystyle y = \frac{180^{\circ }- x}{2}$

$m \angle A = \frac{180^{\circ }- x}{2}$ $\displaystyle m \angle A = \frac{180^{\circ }- x}{2}$

By the same reasoning,

$m \angle D = \frac{180^{\circ }- x}{2}$ $\displaystyle m \angle D = \frac{180^{\circ }- x}{2}$

Therefore, $\angle A \cong \angle D$ $\displaystyle \angle A \cong \angle D$.

By the Angle-Angle Similarity Postulate (AA), two triangles are similar if two angles of the first triangle are congruent to those of their counterparts in the second. $\angle A \cong \angle D$ $\displaystyle \angle A \cong \angle D$ and $\angle B \cong \angle E$ $\displaystyle \angle B \cong \angle E$, so the conditions of AA are met; it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

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Example Question #135 : Acute / Obtuse Triangles

Given: $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$, with Point $\displaystyle M$ on $\overline{AB}$ $\displaystyle \overline{AB}$ and Point $\displaystyle N$ on $\overline{AC}$ $\displaystyle \overline{AC}$.

AM = 10 $\displaystyle AM = 10$

MB = 15 $\displaystyle MB = 15$

AN= 8 $\displaystyle AN= 8$

NC = 16 $\displaystyle NC = 16$

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup AMN$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup AMN$.

Possible Answers:

True

False

Correct answer:

False

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup AMN$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup AMN$, then $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ correspond respectively to $\displaystyle A$, $\displaystyle M$, and $\displaystyle N$.

If two triangles are similar, then it must hold that corresponding sides are in proportion. Specifically, if $\bigtriangleup ABC \sim \bigtriangleup AMN$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup AMN$, it must hold that

$\frac{AB}{AM} = \frac{AC}{AN}$ $\displaystyle \frac{AB}{AM} = \frac{AC}{AN}$

By the Segment Addition Postulate,

$\displaystyle AB = AM + MB = 10+15 = 25$

and

$\displaystyle AC = AN + NC = 8 + 16 = 24$

Set AB = 25 $\displaystyle AB = 25$, AC = 24 $\displaystyle AC = 24$, AM = 10 $\displaystyle AM = 10$, AN = 8 $\displaystyle AN = 8$ in the above proportion statement, which becomes

$\frac{25}{10} = \frac{24}{8}$ $\displaystyle \frac{25}{10} = \frac{24}{8}$

Reduce both ratios to lowest terms:

$\frac{5}{2} = \frac{3}{1}$ $\displaystyle \frac{5}{2} = \frac{3}{1}$

The corresponding sides are not proportional, so the statement $\bigtriangleup ABC \sim \bigtriangleup AMN$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup AMN$ is false.

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Example Question #136 : Acute / Obtuse Triangles

AM = 6 $\displaystyle AM = 6$

MB = 12 $\displaystyle MB = 12$

AN= 9 $\displaystyle AN= 9$

NC = 18 $\displaystyle NC = 18$

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup AMN$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup AMN$.

Possible Answers:

True

False

Correct answer:

True

Explanation:

By the Side-Angle-Side Similarity Theorem (SASS), two triangles are similar if two pairs of corresponding sides are in proportion and their included angles are congruent.

Examine two pairs of corresponding sides: $\overline{AB}$ $\displaystyle \overline{AB}$ and $\overline{AM}$ $\displaystyle \overline{AM}$, and $\overline{AC}$ $\displaystyle \overline{AC}$ and $\overline{AN}$ $\displaystyle \overline{AN}$. In both cases, their included angle is $\angle A$ $\displaystyle \angle A$; by the Reflexive Property of Congruence, $\angle A \cong \angle A$ $\displaystyle \angle A \cong \angle A$.

It remains to be demonstrated that

$\frac{AB}{AM} = \frac{AC}{AN}$ $\displaystyle \frac{AB}{AM} = \frac{AC}{AN}$

By the Segment Addition Postulate,

$\displaystyle AB = AM + MB = 6+12 = 18$

and

$\displaystyle AC = AN + NC = 9+18 = 27$

Set AB = 18 $\displaystyle AB = 18$, AC = 27 $\displaystyle AC = 27$, AM = 6 $\displaystyle AM = 6$, AN= 9 $\displaystyle AN= 9$ in the above proportion statement, which becomes

$\frac{18}{6} = \frac{27}{9}$ $\displaystyle \frac{18}{6} = \frac{27}{9}$

Reduce both ratios to lowest terms:

$\frac{3}{1} = \frac{3}{1}$ $\displaystyle \frac{3}{1} = \frac{3}{1}$

The corresponding sides are proportional.

The conditions of SASS have been proved, so the statement $\bigtriangleup ABC \sim \bigtriangleup AMN$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup AMN$ is true.

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Example Question #137 : Acute / Obtuse Triangles

Given: $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ and $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ such that

$\angle A \cong \angle D$ $\displaystyle \angle A \cong \angle D$

$\angle B \cong \angle E$ $\displaystyle \angle B \cong \angle E$

Which statement(s) must be true?

(a) $\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$

(b) $\bigtriangleup ABC \cong \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF$

Possible Answers:

Neither (a) nor (b)

(b) but not (a)

(a) but not (b)

(a) and (b)

Correct answer:

(a) but not (b)

Explanation:

The two given angle congruences set up the conditions of the Angle-Angle Similarity Postulate - if two angles of one triangle are congruent to the two corresponding angles of another triangle, the two triangles are similar. It follows that

$\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$.

However, congruence cannot be proved, since at least one side congruence is needed to prove this. This is not given in the problem.

Therefore, statement (a) must hold, but not necessarily statement (b).

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Example Question #1 : How To Find The Height Of An Acute / Obtuse Triangle

Inter_geo_tri_series_

Find the height of the triangle shown above.

Possible Answers:

$\displaystyle 16$

$2\sqrt{64}$ $\displaystyle 2\sqrt{64}$

$\displaystyle 4$

$\displaystyle 8$

Correct answer:

$\displaystyle 8$

Explanation:

Use the Pythagorean Theorem to find the height of this triangle: $\displaystyle a^2+b^2=c^2$, where $\displaystyle a=$ the height of the triangle.

$\displaystyle a^2+10^2=164$

$\displaystyle a^2+100=164$

a^2=64 $\displaystyle a^2=64$

$a=\sqrt{64}=8$ $\displaystyle a=\sqrt{64}=8$

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Example Question #2 : How To Find The Height Of An Acute / Obtuse Triangle

Tri_vt

Find the height of the obtuse triangle shown above.

Possible Answers:

$\displaystyle 3$

$\displaystyle 6$

$\displaystyle 4$

$\displaystyle 12$

Correct answer:

$\displaystyle 6$

Explanation:

To solve this solution, first work backwards using the formula:

$Area=\frac{b\cdot h}{2}$ $\displaystyle Area=\frac{b\cdot h}{2}$

Plugging in the given values we are able to solve for the height.
$27=\frac{9h}{2}$ $\displaystyle 27=\frac{9h}{2}$

54= $\displaystyle 54=$$\displaystyle 9h$

$h=\frac{54}{9}=6$ $\displaystyle h=\frac{54}{9}=6$

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Example Question #3 : How To Find The Height Of An Acute / Obtuse Triangle

Tri_vt

Find the height of the acute triangle shown above.

Possible Answers:

$\displaystyle 28$

$\displaystyle 15$

$\displaystyle 20$

$\displaystyle 26$

Correct answer:

$\displaystyle 20$

Explanation:

To solve this solution, work backwards using the formula:

$Area=\frac{b\cdot h}{2}$ $\displaystyle Area=\frac{b\cdot h}{2}$

Plugging in the given values we are able to solve for the height.
$160=\frac{16h}{2}$ $\displaystyle 160=\frac{16h}{2}$

16h=320 $\displaystyle 16h=320$

$h=320\div16=20$ $\displaystyle h=320\div16=20$

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Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #131 : Acute / Obtuse Triangles

Example Question #132 : Acute / Obtuse Triangles

Example Question #133 : Acute / Obtuse Triangles

Example Question #134 : Acute / Obtuse Triangles

Example Question #135 : Acute / Obtuse Triangles

Example Question #136 : Acute / Obtuse Triangles

Example Question #137 : Acute / Obtuse Triangles

Example Question #1 : How To Find The Height Of An Acute / Obtuse Triangle

Example Question #2 : How To Find The Height Of An Acute / Obtuse Triangle

Example Question #3 : How To Find The Height Of An Acute / Obtuse Triangle

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