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PSAT Math : How to find if right triangles are similar

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Example Questions

Example Question #2 :Right Triangles

Triangles

In the figure above, line segmentsDCandABare parallel. What is the perimeter of quadrilateralABCD?

Possible Answers:

Correct answer:

Explanation:

BecauseDCandABare parallel, this means that anglesCDBandABDare equal. When two parallel lines are cut by a transversal line, alternate interior angles (such asCDBandABD) are congruent.

Now, we can show that trianglesABDandBDCare similar. BothABDandBDCare right triangles. This means that they have one angle that is the same—their right angle. Also, we just established that anglesCDBandABDare congruent. By the angle-angle similarity theorem, if two triangles have two angles that are congruent, they are similar. Thus trianglesABDandBDCare similar triangles.

We can use the similarity between trianglesABDandBDCto find the lengths ofBCandCD. The length ofBCis proportional to the length ofAD, and the length ofCDis proportional to the length ofDB, because these sides correspond.

We don’t know the length ofDB, but we can find it using the Pythagorean Theorem. Leta,b, andcrepresent the lengths ofAD,AB, andBDrespectively. According to the Pythagorean Theorem:

a²+b²=c²

15²+ 20²=c²

625 =c²

c= 25

The length ofBDis 25.

Similar_triangles

We now have what we need to find the perimeter of the quadrilateral.

Perimeter = sum of the lengths ofAB,BC,CD, andDA.

Perimeter = 20 + 18.75 + 31.25 + 15 = 85

The answer is 85.

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Example Question #11 :Triangles

$\Delta GHJ \sim \Delta KLM$ and $\angle G$ is a right angle.

Which angle or anglesmustbe complementary to $\angle H$ ?

I) $\angle G$

II) $\angle J$

III) $\angle K$

IV) $\angle L$

V) $\angle M$

Possible Answers:

I only

IV only

II only

II and V only

I and III only

Correct answer:

II and V only

Explanation:

$\angle G$ is a right angle, and, since corresponding angles of similar triangles are congruent, so is $\angle K$ . A right angle cannot be part of a complementary pair so both can be eliminated.

$\angle L$ can be eliminated, since it is congruent to $\angle H$ ; congruent angles are not necessarily complementary.

Since $\angle G$ is right angle, $\Delta GHJ$ is a right triangle, and $\angle H$ and $\angle J$ are its acute angles. That makes $\angle J$ complementary to $\angle H$ . Since $\angle M$ is congruent to $\angle J$ , it is also complementary to $\angle H$ .

The correct response is II and V only.

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Example Question #2 :Right Triangles

Triangles

Refer to the above figure. Given that $\Delta ABC \sim \Delta DE F$ , give the perimeter of $\Delta DE F$ .

Possible Answers:

225

324

135

306

208

Correct answer:

225

Explanation:

By the Pythagorean Theorem,

$BC = \sqrt{(AC)^{2} - (AB)^{2}}$

$= \sqrt{13^{2}- 5^{2}}$

$= \sqrt{169- 25}$

$= \sqrt{144}$

= 12

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{EF}{BC} = \frac{90}{12} = 7.5$ ,

which is subsequently the ratio of the perimeter of $\Delta DE F$ to that of $\Delta ABC$ .

The perimeter of $\Delta ABC$ is

AB + BC + AC = 5 + 12 + 13 = 30 ,

so the perimeter of $\Delta DE F$ can be found using this ratio:

$\frac{\textup{Perim } \Delta DEF}{\textup{Perim } \Delta ABC} = 7.5$

$\frac{\textup{Perim } \Delta DEF}{30} = 7.5$

$\textup{Perim } \Delta DEF = 7.5 \times 30 = 225$

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Example Question #451 :Psat Mathematics

Triangles

Note: Figure NOT drawn to scale.

Refer to the above figure. Given that $\Delta ABC \sim \Delta DE F$ , give the area of $\Delta DE F$ .

Possible Answers:

The correct answer is not among the other responses.

1,828.125

1,687.5

243.75

225

Correct answer:

1,687.5

Explanation:

By the Pythagorean Theorem,

$BC = \sqrt{(AC)^{2} - (AB)^{2}}$

$= \sqrt{13^{2}- 5^{2}}$

$= \sqrt{169- 25}$

$= \sqrt{144}$

= 12

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{EF}{BC} = \frac{90}{12} = 7.5$ ,

This can be used to find:

$\frac{DE}{AB} = 7.5$

$\frac{DE}{5} = 7.5$

$DE = 5 \cdot 7.5 = 37.5$

The area of $\Delta DE F$ is therefore

$\frac{1}{2} \cdot DE \cdot EF = \frac{1}{2} \cdot 37.5 \cdot 90 =1,687.5$

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Example Question #4 :Right Triangles

Triangles

Note: Figures NOT drawn to scale.

Refer to the above figure. Given that $\Delta ABC \sim \Delta DE F$ , evaluate.

Possible Answers:

$52 \frac{1}{2}$

$62\frac{1}{2}$

Correct answer:

Explanation:

By the Pythagorean Theorem, since $\overline{AC}$ is the hypotenuse of a right triangle with legs 6 and 8, its measure is

$AC = \sqrt{6^{2} + 8 ^{2}} = \sqrt{36 + 64 } = \sqrt{100} = 10$ .

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{DF}{AC} = \frac{70}{10} = 7$ .

Likewise,

$\frac{DE}{AB} = 7$

$\frac{DE}{8} = 7$

DE = 56

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PSAT Math : How to find if right triangles are similar

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #2 :Right Triangles

Example Question #11 :Triangles

Example Question #2 :Right Triangles

Example Question #451 :Psat Mathematics

Example Question #4 :Right Triangles

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