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SSAT Upper Level Math : Properties of Triangles

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Example Questions

Example Question #573 : Geometry

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

Possible Answers:

Correct answer:

Explanation:

Add up all the sides to find the perimeter.

2x+4x+5x=121

11x=121

x=11 .

Plugging this value into the sides we get:

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

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

The length of the longest side is .

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

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

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

Possible Answers:

1.25

1.2

1.5

$0.8\overline{3}$

$0.\overline{6}$

Correct answer:

1.5

Explanation:

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

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

Evaluate DE + EF .

Possible Answers:

225

Insufficient information is given to answer the problem.

200

165

210

Correct answer:

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,

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

AB+BC+ AC = 24+ 20+ 36 = 80 , so

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

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

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

DE+EF=165

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

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

Give the perimeter of $\bigtriangleup ABC$ .

Possible Answers:

$53\frac{1}{3}$

360

300

Correct answer:

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,

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

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

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

$P_{\bigtriangleup ABC} = 27$

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

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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

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

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

Possible Answers:

$86^{\circ }$

$188^{\circ }$

These triangles cannot exist.

$172^{\circ }$

$94^{\circ }$

Correct answer:

$172^{\circ }$

Explanation:

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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Example Question #131 : Properties Of Triangles

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 $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

Possible Answers:

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

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

The given information is enough to prove the triangles similar.

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

$\angle B \cong \angle E$

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

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

Possible Answers:

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

$\frac{6}{5}$

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

$\frac{36}{25}$

$\frac{5}{6}$

Correct answer:

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

Explanation:

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

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

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

Possible Answers:

$\bigtriangleup ABC$ is scalene and acute.

None of the other responses is correct.

$\bigtriangleup ABC$ is isosceles and acute.

$\bigtriangleup ABC$ is scalene and obtuse.

$\bigtriangleup ABC$ is isosceles and obtuse.

Correct answer:

$\bigtriangleup ABC$ is scalene and obtuse.

Explanation:

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 $26^{\circ }$ ; its third angle measures

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

One of the angles having measure greater than $90^{\circ }$ 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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Example Question #1 : How To Find If Two Acute / Obtuse Triangles Are Similar

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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, AB < AC < BC .

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

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

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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 #573 : Geometry

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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