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SSAT Upper Level Math : Properties of Triangles

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Example Questions

Example Question #131 : Properties Of Triangles

The lengths of a triangle with a perimeter of 121 $\displaystyle 121$ are $2x, 4x,\text{ and }5x$ $\displaystyle 2x, 4x,\text{ and }5x$. Find the length of the longest side.

Possible Answers:

$\displaystyle 22$

$\displaystyle 55$

$\displaystyle 44$

$\displaystyle 66$

Correct answer:

$\displaystyle 55$

Explanation:

Add up all the sides to find the perimeter.

$\displaystyle 2x+4x+5x=121$

11x=121 $\displaystyle 11x=121$

x=11 $\displaystyle x=11$.

Plugging this value into the sides we get:

$2x, 4x, 5x \rightarrow 2(11), 4(11), 5(11)$ $\displaystyle 2x, 4x, 5x \rightarrow 2(11), 4(11), 5(11)$

The side lengths of the triangle are $22, 44, \text{ and 55}$ $\displaystyle 22, 44, \text{ and 55}$.

The length of the longest side is $\displaystyle 55$.

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Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

$\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$; $\displaystyle AB = 24, BC = 30, AC = 36$; $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ has perimeter 400.

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

Possible Answers:

1.25 $\displaystyle 1.25$

1.2 $\displaystyle 1.2$

1.5 $\displaystyle 1.5$

$0.8\overline{3}$ $\displaystyle 0.8\overline{3}$

$0.\overline{6}$ $\displaystyle 0.\overline{6}$

Correct answer:

1.5 $\displaystyle 1.5$

Explanation:

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

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

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Example Question #2 : How To Find If Two Acute / Obtuse Triangles Are Similar

$\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$; $\displaystyle AB = 24, BC = 20, AC = 36$; $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ has perimeter 300.

Evaluate DE + EF $\displaystyle DE + EF$.

Possible Answers:

225 $\displaystyle 225$

Insufficient information is given to answer the problem.

200 $\displaystyle 200$

165 $\displaystyle 165$

210 $\displaystyle 210$

Correct answer:

165 $\displaystyle 165$

Explanation:

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

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

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

By one of the properties of proportions, it follows that

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

The perimeter of $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is

$\displaystyle AB+BC+ AC = 24+ 20+ 36 = 80$, so

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

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

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

$\displaystyle DE+EF=165$

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Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

$\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$; AB= 12 $\displaystyle AB= 12$; DE = 40 $\displaystyle DE = 40$; $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ has perimeter 90.

Give the perimeter of $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$.

Possible Answers:

$53\frac{1}{3}$ $\displaystyle 53\frac{1}{3}$

$\displaystyle 36$

360 $\displaystyle 360$

$\displaystyle 27$

300 $\displaystyle 300$

Correct answer:

$\displaystyle 27$

Explanation:

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,

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

$\frac{P_{\bigtriangleup ABC }}{90} = \frac{12}{40}$ $\displaystyle \frac{P_{\bigtriangleup ABC }}{90} = \frac{12}{40}$

$\frac{P_{\bigtriangleup ABC }}{90}\cdot 90 = \frac{12}{40} \cdot 90$ $\displaystyle \frac{P_{\bigtriangleup ABC }}{90}\cdot 90 = \frac{12}{40} \cdot 90$

$P_{\bigtriangleup ABC} = 27$ $\displaystyle P_{\bigtriangleup ABC} = 27$

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Example Question #2 : How To Find If Two Acute / Obtuse Triangles Are Similar

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

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

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

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

Possible Answers:

$86^{\circ }$ $\displaystyle 86^{\circ }$

$188^{\circ }$ $\displaystyle 188^{\circ }$

These triangles cannot exist.

$172^{\circ }$ $\displaystyle 172^{\circ }$

$94^{\circ }$ $\displaystyle 94^{\circ }$

Correct answer:

$172^{\circ }$ $\displaystyle 172^{\circ }$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #131 : Properties Of Triangles

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

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

Possible Answers:

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

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

The given information is enough to prove the triangles similar.

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

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

Correct answer:

The given information is enough to prove the triangles similar.

Explanation:

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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Example Question #132 : Properties Of Triangles

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

Possible Answers:

$\frac{625}{1,296}$ $\displaystyle \frac{625}{1,296}$

$\frac{6}{5}$ $\displaystyle \frac{6}{5}$

$\frac{1,296}{625}$ $\displaystyle \frac{1,296}{625}$

$\frac{36}{25}$ $\displaystyle \frac{36}{25}$

$\frac{5}{6}$ $\displaystyle \frac{5}{6}$

Correct answer:

$\frac{1,296}{625}$ $\displaystyle \frac{1,296}{625}$

Explanation:

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

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 $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ to that of $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is

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

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Example Question #5 : How To Find If Two Acute / Obtuse Triangles Are Similar

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

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

Possible Answers:

$\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is scalene and acute.

None of the other responses is correct.

$\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is isosceles and acute.

$\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is scalene and obtuse.

$\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is isosceles and obtuse.

Correct answer:

$\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is scalene and obtuse.

Explanation:

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

Two of the angles of $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ have measures $44 ^{\circ }$ $\displaystyle 44 ^{\circ }$ and $26^{\circ }$ $\displaystyle 26^{\circ }$; its third angle measures

$180^{\circ } - (44^{\circ }+ 26^{\circ } ) = 180^{\circ } - 70 ^{\circ } = 110^{\circ }$ $\displaystyle 180^{\circ } - (44^{\circ }+ 26^{\circ } ) = 180^{\circ } - 70 ^{\circ } = 110^{\circ }$.

One of the angles having measure greater than $90^{\circ }$ $\displaystyle 90^{\circ }$ makes $\bigtriangleup DEF$ $\displaystyle \bigtriangleup DEF$ - and, consequently, $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ - an obtuse triangle. Also, the three angles have different measures, so the sides do as well, making $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ scalene.

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Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

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

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

Possible Answers:

$m \angle C =m \angle B < m \angle A$ $\displaystyle m \angle C =m \angle B < m \angle A$

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

$m \angle C <m \angle B = m \angle A$ $\displaystyle m \angle C < m \angle B = m \angle A$

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

$m \angle B = m \angle C <m \angle A$ $\displaystyle m \angle B = m \angle C < m \angle A$

Correct answer:

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

Explanation:

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

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

Consequently,

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

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

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

Therefore,

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

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Example Question #3 : How To Find If Two Acute / Obtuse Triangles Are Similar

$\bigtriangleup ABC \sim \bigtriangleup DEF$ $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$; $\displaystyle AB = 50, BC = DE= 75, DF = 100$.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

Therefore, $\displaystyle AB < AC < BC$.

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

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

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SSAT Upper Level Math : Properties of Triangles

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #131 : Properties Of Triangles

Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #2 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #2 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #131 : Properties Of Triangles

Example Question #132 : Properties Of Triangles

Example Question #5 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

Example Question #3 : How To Find If Two Acute / Obtuse Triangles Are Similar

All SSAT Upper Level Math Resources

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