How to find if right triangles are similar - PSAT Math

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Question

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

Answer

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:

_a_2 + _b_2 = _c_2

152 + 202 = _c_2

625 = _c_2

c = 25

The length of BD is 25.

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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Question

$\Delta GHJ \sim \Delta KLM$ 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) $\angle M$

Answer

$\angle G$ 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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Question

Note: Figures NOT drawn to scale.

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

Answer

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,

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

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

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Question

Note: Figure NOT drawn to scale.

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

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

Compare your answer with the correct one above

Question

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

,

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

_a_2 + _b_2 = _c_2

152 + 202 = _c_2

625 = _c_2

c = 25

The length of BD is 25.

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.

Compare your answer with the correct one above

Question

$\Delta GHJ \sim \Delta KLM$ 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) $\angle M$

Answer

$\angle G$ 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.

Compare your answer with the correct one above

Question

Note: Figures NOT drawn to scale.

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

Answer

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,

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

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

Compare your answer with the correct one above

Question

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

,

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

_a_2 + _b_2 = _c_2

152 + 202 = _c_2

625 = _c_2

c = 25

The length of BD is 25.

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.

Compare your answer with the correct one above

Question

$\Delta GHJ \sim \Delta KLM$ 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) $\angle M$

Answer

$\angle G$ 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.

Compare your answer with the correct one above

Question

Note: Figures NOT drawn to scale.

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

Answer

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,

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

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

Compare your answer with the correct one above

Question

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

,

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

_a_2 + _b_2 = _c_2

152 + 202 = _c_2

625 = _c_2

c = 25

The length of BD is 25.

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.

Compare your answer with the correct one above

Question

$\Delta GHJ \sim \Delta KLM$ 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) $\angle M$

Answer

$\angle G$ 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.

Compare your answer with the correct one above

Question

Note: Figures NOT drawn to scale.

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

Answer

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,

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

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

Compare your answer with the correct one above

Question

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

Answer

By the Pythagorean Theorem,

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

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

$= \sqrt{169- 25}$

$= \sqrt{144}$

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

,

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

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

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

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

Compare your answer with the correct one above

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