Page 167 - NUMINO Challenge_B1

Answer
Key

2 The length of the drawing paper is 2 times the 2

side of the largest square. Remember that the 4 14 cm
sides of a square are always equal. The sides of 10 cm
each square are found in the order shown
below. Therefore, the perimeter of the shaded 4 cm
square is 5 4 20 cm.
6 cm
15 cm
3 The rectangle is divided into 16 smaller rectangles,
25 cm 10cm 10 cm 5 cm
50 cm 5 cm so the sum of their perimeters is equal to 8 lengths
and 8 widths. The perimeter of the removed
Creative Thinking p.118~p.119 portions is equal to 2 lengths and 2 widths.
Therefore, the perimeter of the remaining 12
1 rectangles is 16 (8 2) 14 (8 2) 180 cm.

4 The perimeter of the remaining space is equal to

(perimeter of the desk) 50 2. The length of the
desk is 40 40 90 170 cm, and the width is
50 80 130 cm. Therefore, the perimeter of the
remaining space is (170 130) 2 100 700 cm.

14 cm

6 cm

2 cm 2 cm 8 cm 14 Measurable Lengths and Weights p.120~p.121
2 cm
Example 7 10, 12 5, 6 11, 11
Each side of Square is 2 cm, each side of
Square is 6 cm and each side of Square is 8 Example 6, 5, 6 1
cm . Each side of Square is 14 cm. Each side of
Square is equal to the sum of the sides of Amount of Water (L)
Squares and . So each side of Square is Steps
8 14 22 cm. Therofore, the perimeter of
Square is 22 4 88 cm. 5-L Container 6-L Container

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NUMINO Challenge B1