Mathematics and Mathematical Literacy - Thutong

Mathematics(NSC)/Grade 10/ P2 82 ExemplarMEMORANDUMGrade 10 Mathematics: Memorandum Paper 21.1.1 A(–2; –5) ⇒ A′(–2; –5) 11.1.2 A(–2; –5) ⇒ A′(2; 5) 11.1.3 A(–2; –5) ⇒ A′(5; –2) 21.2.1AB =22( −3− 5) + (2 − ( −1))1.2.2=73= 8,54 units⎛ − 3 + 5 2 −1⎞M = ⎜ ; ⎟ ⎝ 2 2 ⎠1= ( ; )1 21.2.3 If m BC = 2m BC = p - (-1)2 - 5 = 21.3.11.3.2p + 1 = −6p = −7sin 53,14° = AB20 ∴AB = 20 × sin 53,14°= 16 m tan 53,14° = ABBC AB 16∴BC = = tan 53,14° tan 53,14°21.4.1 V = 18 × 5 × x = 90 x cm³ 21.4.2 New V = 2 × (18 × 5) × x∴new breadth = 2x 11.5.1Mean = 63+32+34+64+32+27+357= 41 21.5.2 Mode = 32 (it occurs most often) 11.5.3 Ages in order: 27; 32; 32; 34; 35; 63; 64 Median = 34 21.5.4 63 22.1223202.3 Solution 1:For ΔABC isosceles to be a right angledisosceles triangle the two equal angles mustbe 45°. In a triangle the longest side isalways opposite the largest angle so inthis triangle the longest side should beopposite the 90°. However, 20 < 5,so ΔABC cannot be a right angledisosceles triangle. Or solution 2:AB = AC,∴if ΔABC is right-angled, Â =90° m AC = 6-2-1-(-4) = 4 3 = 0 2.42.5m ABm AB. m AC ≠ −1 ∴ Â ≠ 90° ∴ ΔABC is NOT right-angled 5⎛ x − 1 y + 6 ⎞ 1E = ⎜ ; ⎟ = (2 ;7)4⎝ 2 2 ⎠x − 1 9 y + 6∴ = and = 7 2 4 29x − 1 = y + 6 = 14211 1x = ( or 5 ) y = 8221∴ D = (5 ;8)236-2m AC =-1-(-4) = 4 3 m AB = 08 − 2 4m BD = = 15 − 1 3m CD =28 − 6=15 − ( −1)2∴ABDC is a trapezium (AC||BD) 53.1.1 Δ2 is the reflection of Δ3 in the y-axis (andvice versa). 23.1.2 Δ1 is the reflection of Δ2 in the x-axis (andvice versa). 23.1.3 Δ2 is the reflection of Δ4 in the line y = x(and vice versa). 23.2.1 Δ3 has been translated 2 units left and 1 unitup. 2413AB = 5 units (sinceBC =m AB = 0 ) 2 2( −1− 1) + (6−2)= 20 32.2 ΔABC isosceles because AB = AC 2Copyright reserved