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SSAT Upper Level Math : How to find if two acute / obtuse triangles are similar

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Example Questions

Example Question #11 : How To Find If Two Acute / Obtuse Triangles Are Similar

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

AB = 20, AC = 30, BC = 25,DE = 50, HI = DF .

Evaluate .

Possible Answers:

100

Correct answer:

Explanation:

Corresponding sides of similar triangles are in proportion, so

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

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

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

DF = 75

Therefore, HI = 75 as well.

Again, by similarity,

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

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

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

GH = 60

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Example Question #142 : Properties Of Triangles

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

$\angle A \cong \angle I$ .

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.

Possible Answers:

II and IV only

I only

I, II, III, and IV

I and III only

II, III, and IV only

Correct answer:

I and III only

Explanation:

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 $90^{\circ }$ , 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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Example Question #143 : Properties Of Triangles

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; AB = 20, BC=DE=40, DF=80 .

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

Possible Answers:

$\bigtriangleup ABC$ is scalene and acute.

$\bigtriangleup ABC$ is isosceles and obtuse.

None of the other responses is correct.

$\bigtriangleup ABC$ is isosceles and acute.

$\bigtriangleup ABC$ is scalene and obtuse.

Correct answer:

$\bigtriangleup ABC$ is isosceles and acute.

Explanation:

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

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

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

AC = 40

Since AC =BC = 40 , $\bigtriangleup ABC$ is isosceles.

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

$20^{2}+40^{2}\; \square \; 40^{2}$

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

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

2,000 > 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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Example Question #144 : Properties Of Triangles

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

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

Possible Answers:

None of the other responses is correct.

$\bigtriangleup DEF$ is isosceles and acute.

$\bigtriangleup DEF$ is scalene and acute.

$\bigtriangleup DEF$ is scalene and obtuse.

$\bigtriangleup DEF$ is isosceles and obtuse.

Correct answer:

$\bigtriangleup DEF$ is isosceles and acute.

Explanation:

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 $90^{\circ }$ , 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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Example Question #581 : Geometry

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; AB = AC = 50, BC = 40 .

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

Possible Answers:

$\bigtriangleup DEF$ is isosceles and obtuse.

$\bigtriangleup DEF$ is scalene and obtuse.

$\bigtriangleup DEF$ is isosceles and acute.

$\bigtriangleup DEF$ is scalene and acute.

None of the other responses is correct.

Correct answer:

$\bigtriangleup DEF$ is isosceles and acute.

Explanation:

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

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

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

DE = DF

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;

$40^{2}+50^{2}\; \square \; 50^{2}$

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

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

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

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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Example Question #142 : Properties Of Triangles

Which of the following statements would prove that the statement

$\bigtriangleup ABC \sim \bigtriangleup DEF$

is false?

Possible Answers:

$m \angle A + m \angle B \ne m \angle D + m \angle E$

$AB + BC \ne DE+EF$

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ have different perimeters

None of the other statements alone would prove the statement $\bigtriangleup ABC \sim \bigtriangleup DEF$ to be false.

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ have different areas

Correct answer:

$m \angle A + m \angle B \ne m \angle D + m \angle E$

Explanation:

Triangles that are similar need not have congruent sides, so it does not follow that AB + BC = DE+EF , 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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Example Question #145 : Properties Of Triangles

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

AB = 30, DE = 40, DF = 60, EF = 50, GI = 150 .

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

Possible Answers:

16:1

100:9

225:2

36:1

625:36

Correct answer:

100:9

Explanation:

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 100:9 .

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Example Question #146 : Properties Of Triangles

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 $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

Possible Answers:

The given information is enough to prove the triangles similar.

$\angle A \cong \angle D$

$\angle C \cong \angle F$

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

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

Correct answer:

$\angle C \cong \angle F$

Explanation:

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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SSAT Upper Level Math : How to find if two acute / obtuse triangles are similar

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #11 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #142 : Properties Of Triangles

Example Question #143 : Properties Of Triangles

Example Question #144 : Properties Of Triangles

Example Question #581 : Geometry

Example Question #142 : Properties Of Triangles

Example Question #145 : Properties Of Triangles

Example Question #146 : Properties Of Triangles

All SSAT Upper Level Math Resources

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