Step 1: Linear Pair & Triangle Sum
\(m\angle PQR = 180^\circ - 125^\circ = 55^\circ\)
\(m\angle R = 180^\circ - (35^\circ + 55^\circ) = 180^\circ - 90^\circ = \mathbf{90^\circ}\) [1 pt]
Class by Sides: Scalene [1.5 pts]
Justification: All interior angles are distinct (\(35^\circ \ne 55^\circ \ne 90^\circ\)), so all three sides must have unequal lengths.
Class by Angles: Right [1.5 pts]
Justification: \(\angle R = 90^\circ\), which constitutes a right angle.
Geometry Assessment • Chapter 4 Answer Key & Scoring Guide Page 1 of 2 See Page 2 for Parts 3, 4, and 5 →
Classifying Triangles Quiz — Answer Key | Page 2: Solutions & Rubric
Parts 3, 4, 5 (20 Points)
Scoring: 1 pt side, 1 pt angle
| Triangle | Classification by SIDES | Classification by ANGLES | Geometric Rationale |
|---|---|---|---|
| #9. \(\triangle ABD\) | Isosceles | Right | \(AB = BD = 10\) (2 tick marks on each leg); \(\angle ABD = 90^\circ\) (marked with square symbol). |
| #10. \(\triangle BCD\) | Scalene | Right | Sides are 6, 10, 12 (all distinct lengths); \(\angle CBD = 90^\circ\) (linear pair with \(\angle ABD\)). |
| #11. \(\triangle ACD\) | Scalene | Obtuse | \(\angle ADC = 45^\circ + 60^\circ = 105^\circ\) (\(>90^\circ\), obtuse); angles are \(45^\circ, 30^\circ, 105^\circ\), so all sides unequal. |
Key Misconception Alert for #11: Students often incorrectly assume \(\triangle ACD\) is acute because vertices \(A\) and \(C\) are acute (\(45^\circ\) and \(30^\circ\)). Emphasize that the vertex angle at \(D\) is the sum of \(\angle ADB\) and \(\angle BDC\): \(45^\circ + 60^\circ = 105^\circ\), which makes the overall triangle obtuse!
Vertices \(J(1,1), K(5,1), L(1,5)\)
a) Side Lengths [3 pts • 1 pt each]
• \(JK = |5 - 1| = \mathbf{4}\text{ units}\) (horizontal distance)
• \(JL = |5 - 1| = \mathbf{4}\text{ units}\) (vertical distance)
• \(KL = \sqrt{(5-1)^2 + (1-5)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = \mathbf{4\sqrt{2}} \approx \mathbf{5.66}\text{ units}\)
b) Slopes & Angle at \(J\) [3 pts • 1 pt each]
• Slope of \(\overline{JK} = \frac{1 - 1}{5 - 1} = \frac{0}{4} = \mathbf{0}\) (horizontal)
• Slope of \(\overline{JL} = \frac{5 - 1}{1 - 1} = \frac{4}{0} =\) Undefined (vertical)
• Angle at \(J\) = 90° (perpendicular axes/lines)
c) Final Precise Dual-Classification [4 pts]:
By Sides: Isosceles (because \(JK = JL = 4\))
By Angles: Right (because \(\angle J = 90^\circ\))
5 pts each
#13. Error Analysis Solution [5 pts]
Verdict: Maya is INCORRECT. [1 pt]
Model Justification [4 pts]:
1. By definition, an equilateral triangle must have three congruent sides, which necessitates that all three interior angles must also be congruent (equiangular).
2. By the Triangle Angle-Sum Theorem, each interior angle must equal \(180^\circ \div 3 = 60^\circ\).
3. A triangle with angles \(90^\circ, 60^\circ, 30^\circ\) has three distinct angle measures, making it a scalene right triangle. In any right triangle, the hypotenuse is strictly longer than each leg (\(c > a\) and \(c > b\)), so all three sides can never be equal.
#14. Classification Matrix Solutions [5 pts]
| Triangle Type | Status | Geometrical Explanation |
|---|---|---|
| Obtuse Equilateral | Impossible | All angles must be \(60^\circ\); cannot have an angle \(>90^\circ\). |
| Right Isosceles | Possible | Angles \(45^\circ - 45^\circ - 90^\circ\); two legs are equal in length. |
| Acute Scalene | Possible | e.g., angles \(50^\circ, 60^\circ, 70^\circ\) (all < 90° and all sides distinct). |
| Obtuse Right | Impossible | \((\text{obtuse} > 90^\circ) + 90^\circ > 180^\circ\), violates angle sum theorem. |
Geometry Assessment • Chapter 4 Answer Key & Scoring Guide Page 2 of 2 Scoring Rubric Complete (40 Points Total)
