Unit Pythagorean Theorem Homework 5 Answer Key

Unit Pythagorean Theorem Homework 5 Answer Key The Pythagorean theorem is a cornerstone of geometry, and mastering its application is essential for success in later math courses. Homework 5 in the unit on the Pythagorean theorem typically presents a mix of computational problems, word‑based scenarios, and proof‑style questions that test both procedural fluency and conceptual understanding. Below you will find a detailed walk‑through of each problem type, the reasoning behind the correct answers, and practical advice to help you avoid common pitfalls.

Introduction

When students encounter unit pythagorean theorem homework 5 answer key, they are looking for more than just a list of numbers; they need a clear explanation of why each step works. This guide breaks down the logic behind the theorem, outlines the structure of Homework 5, and provides a complete, step‑by‑step solution set that you can use to check your work or study for assessments.

Understanding the Pythagorean Theorem

The theorem states that in any right‑angled triangle, the square of the length of the hypotenuse (c) equals the sum of the squares of the lengths of the other two sides (a and b):

[ c^{2}=a^{2}+b^{2} ]

Key points to remember:

• Only right triangles qualify. If a triangle is not right‑angled, the formula does not hold.
• The hypotenuse is always the side opposite the 90° angle and is the longest side.
• The theorem can be rearranged to solve for any missing side:
  • (a = \sqrt{c^{2}-b^{2}})
  • (b = \sqrt{c^{2}-a^{2}}) - (c = \sqrt{a^{2}+b^{2}})

Familiarity with squaring numbers and taking square roots is crucial; many errors arise from arithmetic slips rather than misunderstanding the theorem itself.

Overview of Homework 5

Homework 5 usually contains six to eight problems that progress from basic computation to application. A typical layout looks like this:

Below is a complete answer key with explanations for each problem type, assuming the numbers commonly found in many textbook editions. If your specific homework uses different values, replace the numbers but follow the same reasoning.

Step‑by‑Step Solutions

Problem 1: Find the hypotenuse

Given: legs (a = 6) cm, (b = 8) cm. Solution:

1. Square each leg: (6^{2}=36), (8^{2}=64).
2. Add the squares: (36+64=100).
3. Take the square root: (\sqrt{100}=10).

Answer: (c = 10) cm.

Problem 2: Find the hypotenuse (different numbers)

Given: legs (a = 9) in, (b = 12) in.

Solution:

• (9^{2}=81), (12^{2}=144).
• Sum: (81+144=225).
• (\sqrt{225}=15).

Answer: (c = 15) in.

Problem 3: Find a missing leg

Given: hypotenuse (c = 13) ft, one leg (a = 5) ft. Find (b).

Solution:

1. Square the known quantities: (13^{2}=169), (5^{2}=25). 2. Subtract the leg square from the hypotenuse square: (169-25=144).
2. Square root: (\sqrt{144}=12).

Answer: (b = 12) ft. ---

Problem 4: Find the other leg (alternate)

Given: hypotenuse (c = 26) m, leg (b = 10) m. Find (a).

Solution:

• (26^{2}=676), (10^{2}=100).
• (676-100=576).
• (\sqrt{576}=24).

Answer: (a = 24) m. ---

Problem 5: Verify a right triangle (converse)

Given: side lengths 7, 24, 25.

Solution:

• Identify the longest side as the potential hypotenuse: 25.
• Check: (7^{2}+24^{2}=49+576=625).
• Compare to (25^{2}=625).
• Since the sums are equal, the triangle satisfies the Pythagorean theorem, therefore it is a right triangle.

Answer: Yes, the sides 7‑24‑25 form a right triangle.

Problem 6: Word problem – ladder against a wall

Scenario: A ladder leans against a vertical wall. The foot of the ladder is 3 m from the wall, and the ladder reaches a height of 4 m on the wall. How long is the ladder?

Solution:

1. Sketch a right triangle: wall (vertical leg) = 4 m, ground (horizontal leg) = 3 m, ladder = hypotenuse.
2. Apply

the Pythagorean theorem: (c^{2}=3^{2}+4^{2}).

• Compute the squares: (3^{2}=9) and (4^{2}=16).
• Add them: (9+16=25).
• Take the square root: (\sqrt{25}=5).

Answer: The ladder is 5 m long.

Problem 7: Proof‑style question – show that a triangle with sides 5, 12, 13 is right

Solution:

1. Identify the longest side as the candidate hypotenuse: 13.
2. Square the two shorter sides: (5^{2}=25) and (12^{2}=144).
3. Sum these squares: (25+144=169).
4. Square the longest side: (13^{2}=169).
5. Since (5^{2}+12^{2}=13^{2}), the equality required by the Pythagorean theorem holds.

Conclusion: By the converse of the Pythagorean theorem, a triangle with side lengths 5, 12, 13 is a right triangle.

Problem 8: Challenge problem – coordinate geometry

Scenario: Points (A(2,3)), (B(2,8)), and (C(7,3)) form a triangle. Verify whether (\triangle ABC) is right‑angled and, if so, find its area.

Solution:

1. Compute side lengths using the distance formula (d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}). - (AB): vertical segment, (\sqrt{(2-2)^{2}+(8-3)^{2}}=\sqrt{0+25}=5).

   • (AC): horizontal segment, (\sqrt{(7-2)^{2}+(3-3)^{2}}=\sqrt{25+0}=5).
   • (BC): diagonal, (\sqrt{(7-2)^{2}+(3-8)^{2}}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}).

2. Identify the hypotenuse: The longest side is (BC = 5\sqrt{2}).

3. Apply the Pythagorean check:

   • (AB^{2}+AC^{2}=5^{2}+5^{2}=25+25=50).
   • (BC^{2}=(5\sqrt{2})^{2}=25\cdot2=50).

   Since the sums match, (\triangle ABC) satisfies the theorem and is a right triangle with the right angle at vertex (A).

4. Calculate the area: For a right triangle, area (=\frac{1}{2}\times(\text{leg}{1}\times\text{leg}{2})). - (\text{Area}= \frac{1}{2}\times 5 \times 5 = \frac{25}{2}=12.5) square units. Answer: (\triangle ABC) is right‑angled at (A); its area is (12.5) units². ---

Conclusion

The Pythagorean theorem and its converse provide a powerful toolkit for solving a wide variety of geometric problems—from straightforward calculations of missing sides in right triangles to verifying right‑angle conditions in word problems, proofs, and coordinate‑geometry contexts. By consistently identifying the legs and hypotenuse, squaring the relevant lengths, and comparing sums, students can confidently determine unknown measurements, validate triangle classifications, and even compute areas. Mastery of these steps not only reinforces algebraic manipulation skills but also builds a solid foundation for more advanced topics in trigonometry, vector analysis, and analytic geometry. Continued practice with diverse problem types—such as those outlined above—ensures fluency and readiness for any challenge that relies on the relationship (a^{2}+b^{2}=c^{2}).

Further Explorations ### Generating Pythagorean Triples A Pythagorean triple consists of three positive integers ( (a,b,c) ) that satisfy ( a^{2}+b^{2}=c^{2} ). One classic method to produce infinitely many triples uses two positive integers ( m>n ):

[ a = m^{2}-n^{2},\qquad b = 2mn,\qquad c = m^{2}+n^{2}. ]

For example, choosing ( m=4 ) and ( n=1 ) yields ( (15,8,17) ); checking, ( 15^{2}+8^{2}=225+64=289=17^{2} ). This formula guarantees that every primitive triple (where (a,b,c) share no common factor) appears exactly once when (m) and (n) are coprime and not both odd.

Extending to Three Dimensions

The distance formula in space is a direct analogue of the planar Pythagorean theorem. For points (P(x_{1},y_{1},z_{1})) and (Q(x_{2},y_{2},z_{2})),

[ |PQ|=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}. ]

Consider a rectangular box with side lengths (3), (4), and (12). The space diagonal (d) satisfies [ d^{2}=3^{2}+4^{2}+12^{2}=9+16+144=169;\Longrightarrow;d=13. ]

Thus the triple ( (3,4,12,13) ) illustrates how the theorem generalizes to three‑dimensional right‑angled triangles (the right angle being formed by any two edges meeting at a vertex).

Connection to the Law of Cosines

The law of cosines reduces to the Pythagorean theorem when the included angle is (90^{\circ}). For any triangle with sides (a,b,c) and angle (\gamma) opposite side (c),

[c^{2}=a^{2}+b^{2}-2ab\cos\gamma. ]

If (\gamma=90^{\circ}), (\cos\gamma=0) and the formula collapses to (c^{2}=a^{2}+b^{2}). This perspective shows why the theorem is a special case of a broader relationship and why it appears naturally when dealing with vectors: the dot product of orthogonal vectors is zero, leading to (| \mathbf{u}+\mathbf{v}|^{2}= |\mathbf{u}|^{2}+ |\mathbf{v}|^{2}).

Practice Problems

1. Find the missing leg.
   A right triangle has hypotenuse (26) and one leg (10). Determine the length of the other leg.

2. Verify a triple.
   Show that ( (20,21,29) ) satisfies the Pythagorean relation.

3. 3‑D distance.
   Compute the distance between points (P(1,-2,4)) and (Q(-3,5,-1)).

4. Application.
   A ladder leans against a wall, reaching a height of (8) m. If the base of the ladder is (6) m from the wall, how long is the ladder? *(Solutions can be obtained by applying (a^{2}+b^{2}=c^{

5. Generalize to n dimensions.
   Extend the distance formula to (n) dimensions and discuss how the Pythagorean theorem generalizes in this context.

6. Vector application.
   Given vectors ( \mathbf{u}=(1,2) ) and ( \mathbf{v}=(2,1) ), show that ( |\mathbf{u}+\mathbf{v}|^{2}= |\mathbf{u}|^{2}+ |\mathbf{v}|^{2} ).

7. Historical context.
   Research and discuss the historical development of the Pythagorean theorem, including its origins and early proofs.

8. Cultural significance.
   Explore how the Pythagorean theorem has been used and interpreted across different cultures and time periods.

Conclusion

The Pythagorean theorem, with its elegant simplicity and profound implications, has been a cornerstone of mathematics for millennia. From its roots in ancient geometry to its modern applications in physics and engineering, this theorem continues to inspire and inform. Its ability to generate Pythagorean triples, extend to higher dimensions, and connect with other fundamental mathematical principles underscores its lasting significance. As we continue to explore and apply this theorem, we are reminded of the beauty and power of mathematical relationships that transcend time and culture.