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PSAT Math : How to find if right triangles are similar

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Example Question #1 : Right Triangles

Triangles

In the figure above, line segments DC and AB are parallel. What is the perimeter of quadrilateral ABCD?

Possible Answers:

Correct answer:

Explanation:

Because DC and AB are parallel, this means that angles CDB and ABD are equal. When two parallel lines are cut by a transversal line, alternate interior angles (such as CDB and ABD) are congruent.

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

We can use the similarity between triangles ABD and BDC to find the lengths of BC and CD. The length of BC is proportional to the length of AD, and the length of CD is proportional to the length of DB, because these sides correspond.

We don’t know the length of DB, but we can find it using the Pythagorean Theorem. Let a, b, and c represent the lengths of AD, AB, and BD respectively. According to the Pythagorean Theorem:

a² + b²= c²

15² + 20²= c²

625 = c²

c = 25

The length of BD is 25.

Similar_triangles

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

Perimeter = sum of the lengths of AB, BC, CD, and DA.

Perimeter = 20 + 18.75 + 31.25 + 15 = 85

The answer is 85.

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Example Question #1 : Right Triangles

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

Which angle or angles must be complementary to $\angle H$ ?

I) $\angle G$

II) $\angle J$

III) $\angle K$

IV) $\angle L$

V) $\angle M$

Possible Answers:

II only

I only

II and V only

I and III only

IV 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 #1 : How To Find If Right Triangles Are Similar

Triangles

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

Possible Answers:

135

306

208

225

324

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 #2 : How To Find If Right Triangles Are Similar

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.

1,687.5

225

1,828.125

243.75

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 #2 : How To Find If Right Triangles Are Similar

Triangles

Note: Figures NOT drawn to scale.

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

Possible Answers:

$62\frac{1}{2}$

$52 \frac{1}{2}$

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 #1 : Right Triangles

Example Question #1 : Right Triangles

Example Question #1 : How To Find If Right Triangles Are Similar

Example Question #2 : How To Find If Right Triangles Are Similar

Example Question #2 : How To Find If Right Triangles Are Similar

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