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Intermediate Geometry : How to find if two acute / obtuse triangles are similar

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Example Questions

Example Question #121 : Acute / Obtuse Triangles

Tri_sim_vt_series_cont_

Using the triangle shown above, find possible measurements for the corresponding sides of a similar triangle?

Possible Answers:

and

145 and 165

200 and 213

Correct answer:

145 and 165

Explanation:

Since the two triangles are similar, each triangles three corresponding sides must have the same ratio.

The ratio of the triangle is:

29:33

Applying this ratio we are able to find the lengths of a similar triangle.

$29:33=(29\times 5):(33\times 5)=145:165$

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Example Question #181 : Triangles

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

AB = 10

BC = 15

DE = 12

EF = 18

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

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

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

Possible Answers:

False

True

Correct answer:

True

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

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.

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

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

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

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

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

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

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Example Question #631 : Intermediate Geometry

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

A B = 12

AC = 18

BC = 24

DE = 18

DF = 24

EF = 30

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

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

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

Substituting the lengths of the sides for the respective quantities:

A B = 12

AC = 18

BC = 24

DE = 18

DF = 24

EF = 30

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

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

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 #131 : Acute / Obtuse Triangles

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

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

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

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

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

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

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

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

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

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

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

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

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

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. $m \angle B = 72 ^{\circ }$ and $m \angle E = 72^{\circ }$ , so $\angle B \cong \angle E$ ; $m \angle C = 52 ^{\circ }$ and $m \angle F = 52^{\circ }$ , so $\angle C \cong \angle F$ . The conditions of AA are met, so it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ . The correct response is "true".

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Example Question #132 : Acute / Obtuse Triangles

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

$\angle A \cong \angle C$

$\angle B \cong \angle E$

$\angle D \cong \angle F$

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

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

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

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

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

Substituting:

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

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

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

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

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

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

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

By the same reasoning,

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

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

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. $\angle A \cong \angle D$ and $\angle B \cong \angle E$ , so the conditions of AA are met; it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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Example Question #16 : How To Find If Two Acute / Obtuse Triangles Are Similar

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

AM = 10

MB = 15

AN= 8

NC = 16

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup AMN$ , then , , and correspond respectively to , , and .

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

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

By the Segment Addition Postulate,

AB = AM + MB = 10+15 = 25

and

AC = AN + NC = 8 + 16 = 24

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

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

Reduce both ratios to lowest terms:

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

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

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Example Question #133 : Acute / Obtuse Triangles

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

AM = 6

MB = 12

AN= 9

NC = 18

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup AMN$ , then , , and correspond respectively to , , and .

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

It remains to be demonstrated that

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

By the Segment Addition Postulate,

AB = AM + MB = 6+12 = 18

and

AC = AN + NC = 9+18 = 27

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

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

Reduce both ratios to lowest terms:

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

The corresponding sides are proportional.

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

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Example Question #631 : Plane Geometry

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

$\angle A \cong \angle D$

$\angle B \cong \angle E$

Which statement(s) must be true?

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

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

Possible Answers:

(b) but not (a)

Neither (a) nor (b)

(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

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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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Intermediate Geometry : How to find if two acute / obtuse triangles are similar

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #121 : Acute / Obtuse Triangles

Example Question #181 : Triangles

Example Question #631 : Intermediate Geometry

Example Question #131 : Acute / Obtuse Triangles

Example Question #132 : Acute / Obtuse Triangles

Example Question #16 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #133 : Acute / Obtuse Triangles

Example Question #631 : Plane Geometry

All Intermediate Geometry Resources

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