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Quiz 4-1 Classifying And Solving For Sides/angles In Triangles Answers

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Quiz 4-1 Classifying And Solving For Sides/angles In Triangles Answers
Quiz 4-1 Classifying And Solving For Sides/angles In Triangles Answers

Quiz 4‑1: Classifying and Solving for Sides/Angles in Triangles – Answers and Explanations

In geometry, triangles are the most fundamental shapes, yet they still hold a surprising amount of depth. Below is a comprehensive walk‑through of the questions, the correct answers, and the reasoning behind each step. Quiz 4‑1 focuses on two key skills: (1) classifying triangles by their sides and angles, and (2) solving for unknown sides or angles using the properties of triangles. Whether you’re a student reviewing for an exam or a teacher preparing solutions, this guide will clarify every concept in plain language.


Introduction

Triangles can be sorted in several ways: by side length (equilateral, isosceles, scalene) or by angle measure (acute, right, obtuse). Because of that, once you know how to classify a triangle, you can use the Triangle Sum Theorem (the sum of interior angles is 180°) and the Pythagorean Theorem (for right triangles) to find missing pieces. Quiz 4‑1 tests exactly these skills, and the answers below demonstrate how to approach each problem methodically.


Question 1 – Classify the Triangle

Given: Triangle ABC has side lengths (AB = 7) cm, (BC = 7) cm, and (AC = 10) cm.
Task: Classify by sides and by angles.

Answer

  • By sides: Isosceles – two sides are equal (AB = BC).
  • By angles: Obtuse – the largest side (AC) is opposite the obtuse angle.
    Proof: Apply the Law of Cosines or compare (10^2) to (7^2 + 7^2):
    (10^2 = 100 > 98 = 7^2 + 7^2).
    Therefore the angle opposite AC (∠B) is > 90°.

Why This Works

When two sides are equal, the triangle is isosceles. In practice, to determine the angle type, compare the square of the longest side to the sum of the squares of the other two. If it’s larger, the triangle is obtuse; if smaller, acute; if equal, right.


Question 2 – Find the Missing Angle

Given: Triangle PQR is right-angled at Q. Angles P and R are 35° and 55°, respectively.
Task: Verify the third angle.

Answer

The third angle is 90° by definition of a right triangle. The sum of angles in any triangle is 180°, so
(35° + 55° + 90° = 180°). The given angles are consistent.

Quick Check

If the problem instead gave two acute angles, the third would be (180° - (35° + 55°) = 90°), confirming the right angle.


Question 3 – Solve for a Missing Side in an Isosceles Triangle

Given: Triangle XYZ is isosceles with (XY = XZ = 12) cm. In real terms, the base (YZ) is 10 cm. Find the height from X to base YZ.

Answer

Use the Pythagorean Theorem on the right triangle formed by the height (h), half the base (5) cm, and the equal side (12) cm:

(h^2 + 5^2 = 12^2)
(h^2 + 25 = 144)
(h^2 = 119)
(h = \sqrt{119} \approx 10.90) cm.

Insight

In an isosceles triangle, dropping a perpendicular from the vertex to the base splits the base into two equal segments, creating two congruent right triangles. This symmetry simplifies the calculation.


Question 4 – Use the Law of Sines

Given: Triangle ABC has angles (A = 30°), (B = 45°), and side (a = 8) cm (opposite angle A). Find side (c) (opposite angle C).

Answer

First find angle C: (C = 180° - 30° - 45° = 105°).

Apply the Law of Sines:

[ \frac{a}{\sin A} = \frac{c}{\sin C} ]

[ c = a \cdot \frac{\sin C}{\sin A} = 8 \cdot \frac{\sin 105°}{\sin 30°} ]

[ c \approx 8 \cdot \frac{0.That said, 5} \approx 15. Think about it: 9659}{0. 46\ \text{cm}.

Why Law of Sines?

It relates side lengths to opposite angles in any triangle, not just right triangles. It’s especially useful when two angles and one side are known.


Question 5 – Determine the Type of Triangle from Given Angles

Given: Triangle MNO has angles (M = 90°), (N = 70°), (O = 20°).
Task: Classify by sides.

Answer

Since one angle is 90°, the triangle is a right triangle. To classify by sides, note that the side opposite the 90° angle (MN) is the longest. Even so, without numeric side lengths, we cannot determine if it’s isosceles, scalene, or equilateral. The best we can say is right scalene (assuming typical side lengths; if MN were equal to either other side, it would be right isosceles, but that would require equal angles of 45°, which is not the case).

Takeaway

A right triangle with unequal acute angles is almost always scalene unless given special conditions.


Question 6 – Solve for an Unknown Angle in a Scalene Triangle

Given: Triangle PQR has side lengths (PQ = 9) cm, (QR = 12) cm, (PR = 15) cm. Find angle Q.

Answer

First, identify the longest side: (PR = 15) cm, opposite angle Q. Use the Law of Cosines:

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[ \cos Q = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} ]

[ \cos Q = \frac{9^2 + 12^2 - 15^2}{2 \cdot 9 \cdot 12} ]

[ \cos Q = \frac{81 + 144 - 225}{216} = \frac{0}{216} = 0 ]

So (Q = 90°).

Remark

This triangle is actually a right triangle (a 3‑4‑5 scaled up). Recognizing familiar Pythagorean triples can save time.


Question 7 – Verify Triangle Inequality

Given: Triangle ABC has sides 5 cm, 9 cm, and 12 cm.
Task: Check if these sides can form a triangle.

Answer

Triangle inequality requires that the sum of any two sides exceeds the third:

  1. (5 + 9 = 14 > 12) ✔️
  2. (5 + 12 = 17 > 9) ✔️
  3. (9 + 12 = 21 > 5) ✔️

All conditions are satisfied, so a triangle can be formed.

Quick Test

If any one of these sums were less than or equal to the remaining side, a triangle would not exist.


Question 8 – Find the Height of a Triangle Using Base and Area

Given: Triangle XYZ has a base (YZ = 8) cm and area (A = 18) cm². Find the height (h) from X to YZ.

Answer

Area formula: (A = \frac{1}{2} \times \text{base} \times \text{height}).

[ 18 = \frac{1}{2} \times 8 \times h \implies 18 = 4h \implies h = \frac{18}{4} = 4.5\ \text{cm}. ]

Note

This method works for any triangle when the base and area are known, regardless of the triangle type.


Question 9 – Identify a Triangle as Equilateral

Given: Triangle ABC has all angles equal to 60°.
Task: State the side classification.

Answer

An equilateral triangle has all three sides equal. Since all angles are 60°, the triangle must be equilateral. So, all sides are equal in length.

Confirmation

In an equilateral triangle, each internal angle is (180°/3 = 60°), and the converse is also true: equal angles imply equal sides.


Question 10 – Use the Pythagorean Theorem to Find a Missing Side

Given: Right triangle DEF has legs (DE = 6) cm and (EF = 8) cm. Find the hypotenuse (DF).

Answer

[ DF^2 = DE^2 + EF^2 = 6^2 + 8^2 = 36 + 64 = 100 ]

[ DF = \sqrt{100} = 10\ \text{cm}. ]

Insight

This is another instance of the classic 3‑4‑5 triple scaled by 2. Recognizing such patterns speeds up calculations. Took long enough.


FAQ Section

1. How can I quickly classify a triangle by its angles?

  • Right: One angle is exactly 90°.
  • Acute: All angles < 90°.
  • Obtuse: One angle > 90°.

2. What if I only know two sides of a triangle?

  • Use the Triangle Inequality to confirm a triangle can exist.
  • If one side is opposite a known angle, apply the Law of Sines or Cosines to find missing parts.

3. When is the Law of Sines preferable to the Law of Cosines?

  • When you have two angles and one side (AAS or ASA).
  • When you have two sides and a non‑included angle (SSA), though care is needed due to the ambiguous case.

4. Can I use the Pythagorean Theorem in non‑right triangles?

  • No. The theorem only applies when one angle is exactly 90°. For other triangles, use the Law of Cosines.

5. How do I check if a set of side lengths can form a triangle?

  • Verify the Triangle Inequality for all three combinations.

Conclusion

Quiz 4‑1 covers the essential toolkit for working with triangles: classification, angle calculations, side determinations, and the use of foundational theorems. By mastering these techniques, you gain a solid base for tackling more advanced geometric problems, such as proving properties of triangles, solving for unknowns in composite figures, or working with coordinate geometry. Keep practicing the steps outlined above, and you'll find that the seemingly complex world of triangles becomes intuitive and even enjoyable.

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