GANSBAAI ACADEMIA
MATHEMATICSGrade 10 / /
INVESTIGATION
March 2014
Total: 55Time: 2 x 45 min
EXAMINATOR
MODERATOR / L. Havenga
L. Mostert
memo
TASK 1[13]
Number of triangles / 1 / 2 / 3 / 4 / 7 / 20 / nNumber of matches / 3 / 5 / 7 / 9 / 15 / 41 / 2n+1
Perimeter / 3 / 4 / 5 / 6 / 9 / 22 / n+2
/ / / / /
(½ x 12=6)
a)Double the number of triangles and add 1.(2)
b)Number of matches needed for 25 triangles:
(25 x 2) + 1.= 51 (2)
c)Number of triangles formed with 75 matches:
2n + 1 = 75
=> 2n = 74
=> n = 37(3)
TASK 2[6]
Number of squares / 1 / 2 / 3 / 4 / 7 / 20 / nNumber of matches / 4 / 7 / 10 / 13 / 22 / 41 / 3n+1
Perimeter / 4 / 6 / 8 / 10 / 16 / 42 / 2n+2
/ / / / /
(½ x 12=6)
TASK 3[34]
1. PENTAGON
Broken up, the pattern is:
(2)
Number of pentagons / 1 / 2 / 3 / 4 / 7 / 20 / nNumber of matches / 5 / 9 / 13 / 17 / 29 / 81 / 4n+1
Perimeter / 9 / 8 / 11 / 14 / 23 / 62 / 3n+2
/ / / / / /
(½ x 14=7)
2. HEXAGONS
(2)
Number of hexagons / 1 / 2 / 3 / 4 / 7 / 20 / nNumber of matches / 6 / 11 / 16 / 21 / 36 / 101 / 5n+1
Perimeter / 6 / 10 / 14 / 18 / 30 / 82 / 4n+2
/ / / / / /
(½ x 14=7)
3. OCTAGONS
(2)
Number of octagons / 1 / 2 / 3 / 4 / 7 / 20 / nNumber of matches / 8 / 15 / 22 / 29 / 50 / 141 / 7n+1
Perimeter / 8 / 14 / 20 / 26 / 44 / 122 / 6n+2
/ / / / / /
(½ x 14=7)
c)
Number of sides of polygon / 3 / 4 / 5 / 6 / 8Number of matches for n polygons / 2n+1 / 3n+1 / 4n+1 / 5n+1 / 7n+1
/ / / /
(5)
Number of sides (of polygon) MINUS 1, multiply with number of polygons PLUS 1
(Number of sides – 1) x number of polygons + 1 (2)
TASK 4[2]
OR
OR
(2)
TOTAL: 55
1