Are you struggling to unravel the mysteries of similar triangles? Look no further! In this article, we will provide you with an answer key for Homework 2, paving the way to unlock your understanding of this crucial concept in geometry. With these solutions in hand, you will gain invaluable insights into identifying and applying key properties of similar triangles. Whether you’re a student seeking clarity or an educator looking for a comprehensive resource, our answer key will serve as your ultimate guide to mastering the intricacies of similar triangles. So let’s dive in and unlock the secrets that lie within!

In this article, we will provide the answer key to Homework 2 on the topic of similar triangles. Similar triangles play a crucial role in geometry as they offer opportunities to explore various properties and relationships between shapes. By understanding the key concepts and techniques involved in solving problems related to similar triangles, we can enhance our problem-solving abilities and geometric intuition.

Question 1:

In this question, we are given two triangles, ABC and XYZ. We need to determine whether or not they are similar.

To establish similarity between two triangles, three conditions must be satisfied: AA (Angle-Angle), SAS (Side-Angle-Side), or SSS (Side-Side-Side). Let’s analyze both triangles:

Triangle ABC:

– Angle A is congruent to angle X because they are vertical angles.

– Angle B is congruent to angle Y due to the same reason.

– However, we don’t have enough information about the side lengths of triangle ABC.

Triangle XYZ:

– Angle Y is congruent to angle B due to vertical angles.

– Angle X is congruent to angle A for the same reason.

– Side XY is proportional in length with side BC according to the given proportion XY/BC = 2/5.

Based on this analysis, we can conclude that Triangle ABC and Triangle XYZ are similar by using the AA criterion.

Question 2:

In this question, we have a triangle PQR with two parallel lines intersecting its sides at points S and T. We need to find the lengths of segment ST and QT.

To solve this problem, let’s observe some properties of parallel lines intersecting a triangle:

1. The corresponding angles formed by parallel lines and transversals are congruent.

2. The segments created by these intersections have proportional lengths.

From these observations, we can deduce that triangle PQR is similar to triangle PSR. Therefore, we can set up the proportion PR/PQ = PS/PR.

Using the given information:

PR = 18

PQ = 6

PS = 4

Let’s substitute these values into the proportion:

18/6 = 4/ST

By cross-multiplying, we get:

4 × 6 = 18 × ST

24 = 18ST

Dividing both sides by 18, we find:

ST = 24/18

ST = 4/3

So, the length of segment ST is 4/3.

To find the length of QT, we need to recall another property of parallel lines and transversals: alternate interior angles are congruent. Therefore, angle QTS is congruent to angle QRP.

Since triangle PQR and triangle PSR are similar, their corresponding sides are proportional. We can set up the proportion QR/QP = QS/QR.

Substituting the known values:

QR/PQ = QS/QR

QR/6=4/(4/3)

By cross-multiplying, we obtain:

QR × (4 /3) = 6×4

(4 /3) QR=24

Dividing both sides by (4 /3), we find:

QR=(24/(4 /3))

QR=(72 / 4)

QR=18

Therefore, the length of segment QT is equal to QR which is equal to 18 units.

Summary:

In this article, we provided an answer key for Homework Exercise #2 on similar triangles. By using various criteria such as AA and understanding properties related to parallel lines and transversals in triangles, we were able to solve questions involving similarity concepts effectively. It’s crucial to remember that similarity in triangles implies proportional relationships between corresponding angles and sides. By mastering these concepts and techniques in similar triangles problems, one can strengthen their geometric skills and problem-solving abilities.