Unit 5 Relationships In Triangles Homework 4 Answer Key

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Apr 25, 2025 · 6 min read

Unit 5 Relationships In Triangles Homework 4 Answer Key
Unit 5 Relationships In Triangles Homework 4 Answer Key

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    Unit 5: Relationships in Triangles - Homework 4 Answer Key: A Comprehensive Guide

    This comprehensive guide provides detailed solutions and explanations for Homework 4 in Unit 5: Relationships in Triangles. Understanding these relationships—including similar triangles, triangle congruence postulates, and the properties of medians, altitudes, and angle bisectors—is crucial for mastering geometry. This guide will break down each problem, offering step-by-step solutions and clarifying key concepts. We'll focus on building a strong foundation so you can confidently tackle more advanced geometric problems.

    Understanding the Fundamentals: Key Concepts in Triangle Relationships

    Before diving into the specific problems of Homework 4, let's refresh our understanding of the core concepts:

    1. Similar Triangles:

    Similar triangles have the same shape but not necessarily the same size. This means their corresponding angles are congruent (equal), and their corresponding sides are proportional. Key theorems and postulates related to similar triangles include:

    • AA Similarity Postulate: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
    • SSS Similarity Theorem: If the corresponding sides of two triangles are proportional, then the triangles are similar.
    • SAS Similarity Theorem: If two sides of one triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar.

    2. Congruent Triangles:

    Congruent triangles have the same shape and size. Their corresponding angles and sides are congruent. Key postulates and theorems include:

    • SSS Congruence Postulate: If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.
    • SAS Congruence Postulate: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent.
    • ASA Congruence Postulate: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.
    • AAS Congruence Theorem: If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the triangles are congruent. (Note: HL Congruence Theorem is also relevant, dealing with right-angled triangles.)

    3. Special Segments in Triangles:

    Understanding the properties of medians, altitudes, and angle bisectors is essential.

    • Median: A segment from a vertex to the midpoint of the opposite side. The medians of a triangle intersect at the centroid.
    • Altitude: A perpendicular segment from a vertex to the opposite side (or its extension). The altitudes of a triangle intersect at the orthocenter.
    • Angle Bisector: A segment that divides an angle into two congruent angles. The angle bisectors of a triangle intersect at the incenter.

    Homework 4 Problems and Solutions: A Step-by-Step Approach

    Now, let's tackle the problems from Homework 4. Since I don't have access to your specific homework assignment, I will provide example problems that cover the typical concepts found in a Unit 5: Relationships in Triangles Homework 4 assignment. Remember to adapt these solutions and approaches to your own specific problems.

    Example Problem 1: Similar Triangles

    Problem: In ΔABC, DE || BC. If AD = 4, DB = 6, and AE = 5, find EC.

    Solution:

    Since DE || BC, ΔADE ~ ΔABC by AA Similarity (corresponding angles are congruent). Therefore, the ratios of corresponding sides are equal:

    AD/AB = AE/AC

    Substitute the given values:

    4/(4+6) = 5/AC

    4/10 = 5/AC

    Solving for AC:

    AC = (10 * 5) / 4 = 12.5

    Therefore, EC = AC - AE = 12.5 - 5 = 7.5

    Example Problem 2: Congruent Triangles

    Problem: Given ΔABC and ΔDEF, AB = DE, BC = EF, and ∠B = ∠E. Prove that ΔABC ≅ ΔDEF.

    Solution:

    We can use the SAS (Side-Angle-Side) Congruence Postulate to prove the triangles are congruent. We are given:

    • AB = DE (Side)
    • BC = EF (Side)
    • ∠B = ∠E (Angle)

    Since the included angle (∠B in ΔABC and ∠E in ΔDEF) is congruent, and the two adjacent sides are congruent, the SAS Congruence Postulate applies, proving ΔABC ≅ ΔDEF.

    Example Problem 3: Medians and Centroids

    Problem: The medians of a triangle intersect at the centroid. If the distance from a vertex to the centroid is 12, what is the length of the median from that vertex?

    Solution:

    The centroid divides each median into a 2:1 ratio. The longer segment (from the vertex to the centroid) is twice the length of the shorter segment (from the centroid to the midpoint of the opposite side).

    Let x be the length of the shorter segment. Then the length of the longer segment is 2x.

    We are given that the longer segment (vertex to centroid) is 12, so 2x = 12. Solving for x, we get x = 6.

    The length of the median is the sum of the two segments: 12 + 6 = 18.

    Example Problem 4: Altitudes and Orthocenter

    Problem: In a right-angled triangle, where is the orthocenter located?

    Solution:

    In a right-angled triangle, the orthocenter is located at the vertex containing the right angle.

    Example Problem 5: Angle Bisectors and Incenter

    Problem: An angle bisector of a triangle divides the opposite side into segments with lengths proportional to the lengths of the adjacent sides. If an angle bisector divides a side into segments of length 6 and 8, and one adjacent side has length 10, what is the length of the other adjacent side?

    Solution:

    Let the lengths of the sides be a, b, and c. The angle bisector divides the opposite side into segments of lengths m and n. The Angle Bisector Theorem states:

    a/b = m/n

    We are given m = 6, n = 8, and one adjacent side (a) = 10. We need to find b:

    10/b = 6/8

    Solving for b:

    b = (10 * 8) / 6 = 40/3

    Expanding Your Understanding: Further Exploration of Triangle Relationships

    To further solidify your understanding, consider exploring these advanced topics:

    • Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
    • Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
    • Trigonometric Ratios: Sine, cosine, and tangent relate the angles and sides of a right-angled triangle.
    • Area formulas for triangles: Heron's formula, and the formula using base and height.

    By consistently practicing and applying these concepts, you'll build a strong foundation in geometry and excel in future mathematical endeavors. Remember to review your class notes, consult your textbook, and seek help from your teacher or tutor if you encounter difficulties. Good luck!

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