Page 166 - NUMINO Challenge_B1
P. 166
Measurement Problem solving
1 If the width of the rectangle is cm, then the length
13 Perimeter p.112~p.113 of the rectangle is 4 cm, so the perimeter of one
rectangle is 2 ( 4 ) 10 cm.
Example 36 20 16, 16, 48 104 Since 10 30 cm, we obtain 3 cm. The side
of the square is 3 4 12 cm, so the perimeter of
the original square is 12 4 48 cm.
Example 24 cm 9 cm 2 When the rectangle is cut horizontally three
15 cm 15 cm
times and vertically twice, 6 sides with the same
3, 12 9 cm length and 4 sides with the same width are
6 cm 3 cm created. Therefore, the sum of the perimeters of
the 12 rectangles is
6 cm 3 cm (12 10) 2 12 6 10 4 156 cm.
3 cm
Type 13-2 Perimeter of a Combined Shape p.116~p.117
Type 13-1 Sum of the Perimeters p.114~p.115 1
1 100 4 25 cm 2 The sum of the blue sides is 4 4 16 cm. The
sum of the green sides is equal to 3 times the
2 The length of each rectangle is equal to a fifth side of the square, so it is 4 3 12 cm.
of the side of the square, so each rectangle is
25 5 5 cm long. 3 The sum of the remaining sides is 4 7 28 cm.
Therefore, the perimeter of the shape is 16
3 The width of each rectangle is equal to the side 12 28 56 cm.
of the square, so it is 25 cm wide.
Problem solving
4 (25 5) 2 60 cm
1
5 Each side of the square is 60 4 15 cm, so the
length of the rectangle is 15 3 5 cm and the
width is 15 cm. Therefore, the perimeter of each
rectangle is (5 15) 2 40 cm.
The perimeter of the shape is equal to 18 times
the side of a square. Therefore, the perimeter is
3 18 54 cm.
Answer Key
1 If the width of the rectangle is cm, then the length
13 Perimeter p.112~p.113 of the rectangle is 4 cm, so the perimeter of one
rectangle is 2 ( 4 ) 10 cm.
Example 36 20 16, 16, 48 104 Since 10 30 cm, we obtain 3 cm. The side
of the square is 3 4 12 cm, so the perimeter of
the original square is 12 4 48 cm.
Example 24 cm 9 cm 2 When the rectangle is cut horizontally three
15 cm 15 cm
times and vertically twice, 6 sides with the same
3, 12 9 cm length and 4 sides with the same width are
6 cm 3 cm created. Therefore, the sum of the perimeters of
the 12 rectangles is
6 cm 3 cm (12 10) 2 12 6 10 4 156 cm.
3 cm
Type 13-2 Perimeter of a Combined Shape p.116~p.117
Type 13-1 Sum of the Perimeters p.114~p.115 1
1 100 4 25 cm 2 The sum of the blue sides is 4 4 16 cm. The
sum of the green sides is equal to 3 times the
2 The length of each rectangle is equal to a fifth side of the square, so it is 4 3 12 cm.
of the side of the square, so each rectangle is
25 5 5 cm long. 3 The sum of the remaining sides is 4 7 28 cm.
Therefore, the perimeter of the shape is 16
3 The width of each rectangle is equal to the side 12 28 56 cm.
of the square, so it is 25 cm wide.
Problem solving
4 (25 5) 2 60 cm
1
5 Each side of the square is 60 4 15 cm, so the
length of the rectangle is 15 3 5 cm and the
width is 15 cm. Therefore, the perimeter of each
rectangle is (5 15) 2 40 cm.
The perimeter of the shape is equal to 18 times
the side of a square. Therefore, the perimeter is
3 18 54 cm.
Answer Key
