How to find if two acute / obtuse triangles are similar - SSAT Upper Level Quantitative

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; $\bigtriangleup DEF$ has perimeter 400.

Which of the following is equal to $DF \div DE$ ?

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Answer

The perimeter of $\bigtriangleup DEF$ is actually irrelevant to this problem. Corresponding sides of similar triangles are in proportion, so use this to calculate $DF \div DE$ , or $\frac{DF}{DE}$ :

$$\frac{DF}{DE}$ = $\frac{AC}{AB}$ = $\frac{36}{24}$ = 1.5$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; $\bigtriangleup DEF$ has perimeter 300.

Evaluate .

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Answer

The ratio of the perimeters of two similar triangles is equal to the ratio of the lengths of a pair of corresponding sides. Therefore,

$$\frac{DE}{AB}$= $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$ and $$\frac{EF}{BC}$= $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$ , or

$$\frac{DE}{AB}$=\frac{EF}{BC}$ = $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$

By one of the properties of proportions, it follows that

$$\frac{DE+EF}{AB+BC}$ = $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$

The perimeter of $\bigtriangleup ABC$ is

, so

$\frac{DE+EF}{24+20}$ = $\frac{300}{80}$

$\frac{DE+EF}{44}$ = $\frac{300}{80}$

$$\frac{DE+EF}{44}$ \cdot 44 = $\frac{300}{80}$ \cdot 44$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; ; $\bigtriangleup DEF$ has perimeter 90.

Give the perimeter of $\bigtriangleup ABC$ .

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Answer

The ratio of the perimeters of two similar triangles is the same as the ratio of the lengths of a pair of corresponding sides. Therefore,

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

$m\angle A + m\angle D = $94^{\circ }$$

$m\angle B + m\angle F = $94^{\circ }$$

Evaluate $m \angle C + m \angle E$ .

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Answer

The similarity of the triangles is actually extraneous information here. The sum of the measures of a triangle is $180 ^{\circ }$ , so:

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

$m \angle D + m \angle E + m \angle F = $180^{\circ }$$

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

$m \angle C + m \angle E + (m \angle A +m \angle D) + (m \angle B + m \angle F )= $360^{\circ }$$

$m \angle C + m \angle E +94 ^{\circ } +94 ^{\circ }= $360^{\circ }$$

$m \angle C + m \angle E +188 ^{\circ }= $360^{\circ }$$

$m \angle C + m \angle E +188 ^{\circ }- $188^{\circ }$ = $360^{\circ }$ - $188^{\circ }$$

$m \angle C + m \angle E = $172^{\circ }$$

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ ; $\frac{AB}{DE}$ = $\frac{AC}{DF}$ .

Which of the following statements would not be enough, along with what is given, to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

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Answer

From both the given proportion statement and either $\frac{AB}{DE}$ = $\frac{BC}{EF}$ or $\frac{AC}{DF}$= $\frac{BC}{EF}$ , it follows that $\frac{AB}{DE}$ =\frac{AC}{DF}$= $\frac{BC}{EF}$ —all three pairs of corresponding sides are in proportion; by the Side-Side-Side Similarity Theorem, $\bigtriangleup ABC \sim \bigtriangleup DEF$ . From the given proportion statement and $\angle A \cong \angle D$ , since these are the included angles of the sides that are in proportion, then by the Side-Angle-Side Similarity Theorem, $\bigtriangleup ABC \sim \bigtriangleup DEF$ . From the given proportion statement and $\angle C \cong \angle F$ , since these are nonincluded angles of the sides that are in proportion, no similarity can be deduced.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ ; $\angle A \cong \angle D$ and $\angle C \cong \angle F$ .

Which of the following statements would not be enough, along with what is given, to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

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Answer

Two pairs of corresponding angles are stated to be congruent in the main body of the problem; it follows from the Angle-Angle Similarity Postulate that the triangles are similar. No further information is needed.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; .

Which of the following is true about $\bigtriangleup DEF$ ?

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Answer

$\bigtriangleup ABC \sim \bigtriangleup DEF$ , so corresponding sides are in proportion; it follows that

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

$\frac{DE}{50}$ = $\frac{DF}{50 }$

Therefore, $\bigtriangleup DEF$ is isosceles.

Also, corresponding angles are congruent, so if $\bigtriangleup ABC$ acute (or obtuse), so is $\bigtriangleup DEF$ . We can compare the sum of the squares of the lesser two sides to that of the greatest;

$1,600+2,500; \square ; 2,500$

$4,100; \square ; 2,500$

The sum of the squares of the lesser sides is greater than the square of the greatest side, so $\bigtriangleup ABC$ is acute - and so is $\bigtriangleup DEF$ . The correct response is that $\bigtriangleup DEF$ is isosceles and acute.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $m \angle D = 44 ^{\circ }, m \angle E = $26^{\circ }$$

Which of the following is true about $\bigtriangleup ABC$ ?

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Answer

Corresponding angles of similar triangles are congruent, so the measures of the angles of $\bigtriangleup ABC$ are equal to those of $\bigtriangleup DEF$ .

Two of the angles of $\bigtriangleup DEF$ have measures $44 ^{\circ }$ and ; its third angle measures

.

One of the angles having measure greater than makes $\bigtriangleup DEF$ - and, consequently, $\bigtriangleup ABC$ - an obtuse triangle. Also, the three angles have different measures, so the sides do as well, making $\bigtriangleup ABC$ scalene.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ . $\frac{AB}{DE}$ = $\frac{25}{36}$ . Which of the following is the ratio of the area of $\bigtriangleup DEF$ to that of $\bigtriangleup ABC$ ?

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Answer

The similarity ratio of $\bigtriangleup ABC$ to $\bigtriangleup DEF$ is equal to the ratio of two corresponding sidelengths, which is given as $\frac{25}{36}$ ; the similarity ratio of $\bigtriangleup DEF$ to $\bigtriangleup ABC$ is the reciprocal of this, or $\frac{36}{25}$ .

The ratio of the area of a figure to that of one to which it is similar is the square of the similarity ratio, so the ratio of the area of $\bigtriangleup DEF$ to that of $\bigtriangleup ABC$ is

$\left ($\frac{36}{25}$ \right ) ^{2}= $\frac{1,296}{625}$$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $m \angle A = 74 ^{\circ }, m \angle E = $32^{\circ }$$

Which of the following is true about $\bigtriangleup DEF$ ?

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Answer

Corresponding angles of similar triangles are congruent, so the measures of the angles of $\bigtriangleup ABC$ are equal to those of $\bigtriangleup DEF$ .

$\angle D \cong \angle A$ , so $m \angle D = m \angle A = 74 ^{\circ }$ . Also $m \angle E = $32^{\circ }$$ , so

$m \angle F = $180^{\circ }$ - (m \angle D + m \angle E )$ .

$= $180^{\circ }$ - $(74^{\circ }$+ $32^{\circ }$ )$

$= $180^{\circ }$ - $106^{\circ }$$

$= 74 ^{\circ }$

All three angles have measure less than , so $\bigtriangleup DEF$ is acute. Also, two of the angles are congruent, so by the Converse of the Isosceles Triangle Theorem, $\bigtriangleup DEF$ is isosceles.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; .

Which of the following is true about $\bigtriangleup ABC$ ?

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Answer

Corresponding sides of similar triangles are in proportion, so

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

Substituting, we have from the first statement

$\frac{AC}{80}$ = $\frac{20}{40}$

$$\frac{AC}{80}$ \cdot 80 = $\frac{20}{40}$ \cdot 80$

Since , $\bigtriangleup ABC$ is isosceles.

We can compare the sum of the squares of the lesser two sides to that of the greatest.

$400+ 1,600 ; \square ; 1,600$

$2,000 ; \square ; 1,600$

The sum of the squares of the lesser two sides is greater than the square of the third, so $\bigtriangleup ABC$ is acute.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $m \angle A = $66^{\circ }$$ ; $m \angle F = $63^{\circ }$$ .

Which of the following correctly gives the relationship of the angles of $\bigtriangleup ABC$ ?

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Answer

$m \angle A = $66^{\circ }$$

Corresponding angles of similar triangles are congruent, so $m \angle C =m \angle F = $63^{\circ }$$ .

Consequently,

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

$m \angle B = $180^{\circ }$ - ( $66^{\circ }$+ $63^{\circ }$)$

$m \angle B = $180^{\circ }$ - $129^{\circ }$ = $51^{\circ }$$

Therefore,

$m \angle B < m \angle C <m \angle A$ .

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Question

Which of the following statements would prove that the statement

$\bigtriangleup ABC \sim \bigtriangleup DEF$

is false?

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Answer

Triangles that are similar need not have congruent sides, so it does not follow that , or that their perimeters are equal. Consequently, their areas need not be equal either.

However, if $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then corresponding angles are congruent; specifically, $m \angle A = m \angle D$ and $m \angle B =m \angle E$ . Therefore, $m \angle A + m \angle B = m \angle D + m \angle E$ . Contrapositively, if $m \angle A + m \angle B \ne m \angle D + m \angle E$ , then $\bigtriangleup ABC \nsim \bigtriangleup DEF$ .

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF \sim \bigtriangleup GHI$ .

.

Evaluate .

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Answer

Corresponding sides of similar triangles are in proportion, so

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

$\frac{DF}{30}$ = $\frac{50}{20}$

$$\frac{DF}{30}$ \cdot 30 = $\frac{50}{20}$\cdot 30$

Therefore, as well.

Again, by similarity,

$\frac{GH}{AB}$ = $\frac{HI}{BC}$

$\frac{GH}{20}$ = $\frac{75}{25}$

$$\frac{GH}{20}$\cdot 20 = $\frac{75}{25}$ \cdot 20$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; .

Which of the following correctly gives the relationship of the angles of $\bigtriangleup ABC$ ?

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Answer

Corresponding sides of similar triangles are in proportion; since $\bigtriangleup ABC \sim \bigtriangleup DEF$ ,

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

$\frac{AC}{100}$ = $\frac{50}{75}$

$$\frac{AC}{100}$ \cdot 100 = $\frac{50}{75}$ \cdot 100$

$AC = 66$\frac{2}{3}$$

Therefore, .

The angle opposite the longest (shortest) side of a triangle is the angle of greatest (least) measure, so

$m \angle C <m \angle B < m \angle A$ .

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF \sim \bigtriangleup GHI$ .

$\angle A \cong \angle I$ .

Which of the following scenarios is possible ?

I) $\bigtriangleup DEF$ is an acute triangle.

II) $\bigtriangleup DEF$ is an obtuse triangle with $\angle D$ the obtuse angle.

III) $\bigtriangleup DEF$ is an obtuse triangle with $\angle E$ the obtuse angle.

IV) $\bigtriangleup DEF$ is an obtuse triangle with $\angle F$ the obtuse angle.

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Answer

Corresponding angles of similar triangles are congruent, so $\angle A \cong \angle D$ and $\angle F \cong \angle I$ . Since $\angle A \cong \angle I$ , it follows that $\angle D \cong \angle F$ . Since a triangle cannot have two angles that measure more than , both $\angle D$ and $\angle F$ are acute. No information is given about $\angle E$ , so $\angle E$ can be acute, right, or obtuse. Therefore, scenarios (I) and (III) are possible, but not (II) or (IV).

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF \sim \bigtriangleup GHI$ .

Order the triangles by perimeter, least to greatest.

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Answer

$\overline{DF}$ and $\overline{AC}$ are corresponding sides of their respective triangles, and , so it easily follows from proportionality that each side of $\bigtriangleup DEF$ is shorter than its corresponding side in $\bigtriangleup ABC$ . Therefore, $\bigtriangleup DEF$ is of lesser perimeter than $\bigtriangleup ABC$ . By the same reasoning, since , $\bigtriangleup ABC$ is of lesser perimeter than $\bigtriangleup GHI$ .

The correct response is

$\bigtriangleup DEF, \bigtriangleup ABC,\bigtriangleup GHI$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF \sim \bigtriangleup GHI$ .

.

What is the ratio of the area of $\bigtriangleup GHI$ to that of $\bigtriangleup ABC$ ?

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Answer

The similarity ratio of two triangles is the ratio of the lengths of their corresponding sides.

The similarity ratio of $\bigtriangleup DEF$ to $\bigtriangleup ABC$ is

$\frac{DE}{AB}$ = $\frac{40}{30}$ = $\frac{4}{3}$ .

The similarity ratio of $\bigtriangleup GHI$ to $\bigtriangleup DEF$ is

$\frac{GI}{DF}$ = $\frac{150}{60}$ = $\frac{5}{2}$

Multipliy these to get the similarity ratio of $\bigtriangleup GHI$ to $\bigtriangleup ABC$ :

$\frac{4}{3}$ \times $\frac{5}{2}$ = $\frac{20}{6}$= $\frac{10}{3}$

The ratio of the areas of two similar figures is the square of their similarity ratio, so the ratio of the areas of the triangles is

$\left ( $\frac{10}{3}$ \right $)^{2}$ = $\frac{100}{9}$$ .

The correct choice is .

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; $\bigtriangleup DEF$ has perimeter 400.

Which of the following is equal to $DF \div DE$ ?

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Answer

The perimeter of $\bigtriangleup DEF$ is actually irrelevant to this problem. Corresponding sides of similar triangles are in proportion, so use this to calculate $DF \div DE$ , or $\frac{DF}{DE}$ :

$$\frac{DF}{DE}$ = $\frac{AC}{AB}$ = $\frac{36}{24}$ = 1.5$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; ; $\bigtriangleup DEF$ has perimeter 300.

Evaluate .

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Answer

The ratio of the perimeters of two similar triangles is equal to the ratio of the lengths of a pair of corresponding sides. Therefore,

$$\frac{DE}{AB}$= $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$ and $$\frac{EF}{BC}$= $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$ , or

$$\frac{DE}{AB}$=\frac{EF}{BC}$ = $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$

By one of the properties of proportions, it follows that

$$\frac{DE+EF}{AB+BC}$ = $$\frac{P_{\bigtriangleup DEF}$$$}{P_{\bigtriangleup ABC }$}$

The perimeter of $\bigtriangleup ABC$ is

, so

$\frac{DE+EF}{24+20}$ = $\frac{300}{80}$

$\frac{DE+EF}{44}$ = $\frac{300}{80}$

$$\frac{DE+EF}{44}$ \cdot 44 = $\frac{300}{80}$ \cdot 44$

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