Reasoning and Proof: math test solutions, Grade 10 – download the PDF

1 Conditional statements / 4 pts

1. Converse: “If a quadrilateral has four right angles, then it is a rectangle.” True (1 pt). Inverse: “If a quadrilateral is not a rectangle, then it does not have four right angles.” True (1 pt). Contrapositive: “If a quadrilateral does not have four right angles, then it is not a rectangle.” True (1 pt).
2. Because the statement and its converse are both true: “A quadrilateral is a rectangle if and only if it has four right angles.” (1 pt)

2 Reasoning and counterexamples / 3 pts

1. Inductive reasoning (1 pt).
2. The number 20 is a multiple of 5 and is even (1 pt).
3. \(3+5=8\), and 3 and 5 are prime but 8 is not (1 pt).

3 Algebra proof / 4 pts

1. \(4(x+3)=2x+30\) (Given). 2. \(4x+12=2x+30\) (Distributive Property) (1 pt). 3. \(2x+12=30\) (Subtraction Property: subtract \(2x\)) (1 pt). 4. \(2x=18\) (Subtraction Property: subtract 12) (1 pt). 5. \(x=9\) (Division Property) (1 pt).

Check: \(4\cdot 12=48=2\cdot 9+30\).

4 Angle addition / 4 pts

Angle Addition Postulate (1 pt): \((4x-6)+(2x+12)=108\) (1 pt). \(6x+6=108\), so \(x=17\) (1 pt). \(m\angle ABD=4\cdot 17-6=62^\circ\) and \(m\angle DBC=2\cdot 17+12=46^\circ\); \(62+46=108\) (1 pt).

5 Paragraph and flow proof / 5 pts

1. Since \(\angle 1\) and \(\angle 3\) are complementary, \(m\angle 1+m\angle 3=90^\circ\) (1 pt). Since \(\angle 2\) and \(\angle 3\) are complementary, \(m\angle 2+m\angle 3=90^\circ\) (1 pt). By substitution, \(m\angle 1+m\angle 3=m\angle 2+m\angle 3\), and subtracting \(m\angle 3\) gives \(m\angle 1=m\angle 2\), so \(\angle 1\cong\angle 2\) (1 pt).
2. Box 1: \(m\angle 1+m\angle 3=90^\circ\) (Definition of complementary angles). Box 2: \(m\angle 2+m\angle 3=90^\circ\) (Definition of complementary angles). Box 3: \(m\angle 1+m\angle 3=m\angle 2+m\angle 3\) (Substitution), with arrows from boxes 1 and 2. Box 4: \(m\angle 1=m\angle 2\) (Subtraction Property), then \(\angle 1\cong\angle 2\) (Definition of congruent angles) (2 pts).