Page 154 - NUMINO Challenge_C2

Problem solving 3

1 According to Pick's Theorem, 4 2 1 1; { 6 } { 7 } { 9 }
3 3 4
therefore, there are four dots on the perimeter
of the shape and no dots inside the shape.

To have the shapes connected as those in { 5 },
2

{ 7 }, { 7 }, and { 9 }, the common divisor of
2 3 4

and should only be 1.

2 The area of the shape on the dotted board 8 Combining and Splitting Shapes p.68~p.69

increases as both the number of dots on the

perimeter of the shape and the number of dots

inside the shape increase.

Number of Dots on the Number of Dots Example 1 2 4
Perimeter of the Shape in the Interior
Example 6, 24 6
8 3
8 5
9 5

The shape has the greatest area.

Creative Thinking p.66~p.67 Type 8-1 Combining Different Shapes p.70~p.71
1
1 1234 1234 1234 1234 2

12 11 10 9 12 11 10 9 12 11 10 9 12 11 10 9 Answer Key

8 765 8 765 8765 8765

13 14 15 16 13 14 15 16 13 14 15 16 13 14 15 16

Target Number: 34

2 If there is no hole, since there are seven dots on

the perimeter of shape A and six dots in the
interior, the area of shape A is 7 2 6 1

8.5. Also, since the area of is 1; therefore,

the area of is 2. Therefore, the area of

shape A is 8.5 2 6.5.
If there is no hole, since there are 10 dots on the
perimeter of shape B and eight dots in the
interior, the area of shape B is 10 2 8 1

12. Also, since the area of is 1, then the

area of and the area of is 1. Therefore,

the area of shape B is 12 1.5 10.5.