Page 156 - NUMINO Challenge_D1
P. 156
Problem solving Creative Thinking p.74~p.75
1 The base is a square and the side faces are 1 The lower solid figure is the frustum of a
isosceles triangles, so it is a rectangular pyramid. pyramid. If the number of edges of the base is
The side of the square is 3 cm and the base and , the number of vertices of the frustum is
height of the triangle are 3 cm and 6 cm
respectively. Therefore, the area is 3 3 (3 6 2) 2. Therefore, the base of the frustum of the
pyramid has 13 sides. The upper solid figure is a
4 45 cm2. pyramid with a base of 13 sides. The number of
faces in the lower figure is 15 and the number of
2 The solid figure is an octagonal prism. faces in the upper figure is 14. Therefore, the
sum is 14 15 29.
3 cm
2 A is a prism and B is a pyramid. A prism with a
8 cm
base of sides has 3 edges. A pyramid
(3 8) 2 8 8 112 cm
with a base of sides has 2 edges. A
common multiple of 3 and 2 is 6. Draw a table
with the number of edges as multiples of 6.
Type 8-2 Dual Polyhedron p.72~p.73 Edges 6 12 18 24
1
Sides of the impossible 4 6 8
base of prism
Vertices of impossible 8 8 16
prism
Sides of the 3 6 9 12
base of pyramid
2 Regular octahedron Vertices of 4 7 10 13
pyramid
Difference in 123
vertices
3 Faces Vertices Edges
Cube 6 8 12 Therefore, A is a rectangular prism because it
8 6 12 has a rectangular base, and B is a hexagonal
pyramid because it has a hexagonal base.
4 Regular octahedron, vertices 3 At least three faces must meet at one vertex to
Problem solving make a solid figure. Remember that the angle of
an equilateral triangle is 60 . So if three, four, or
1 The dual of a regular tetrahedron is a regular five equilateral triangles meet at one vertex,
each interior angle becomes 180 , 240 , or
tetrahedron. 300 , respectively. It is impossible for more than
six faces to meet at a vertex because each
2 The dual of a regular octahedron is a cube. interior angle would be greater than 360 , which
is impossible. Therefore, only three regular
There are 6 faces and 8 vertices in a cube. polyhedrons: the tetrahedron, octahedron, and
Therefore, the sum is 6 8 14. icosahedron can be made.
Answer Key
1 The base is a square and the side faces are 1 The lower solid figure is the frustum of a
isosceles triangles, so it is a rectangular pyramid. pyramid. If the number of edges of the base is
The side of the square is 3 cm and the base and , the number of vertices of the frustum is
height of the triangle are 3 cm and 6 cm
respectively. Therefore, the area is 3 3 (3 6 2) 2. Therefore, the base of the frustum of the
pyramid has 13 sides. The upper solid figure is a
4 45 cm2. pyramid with a base of 13 sides. The number of
faces in the lower figure is 15 and the number of
2 The solid figure is an octagonal prism. faces in the upper figure is 14. Therefore, the
sum is 14 15 29.
3 cm
2 A is a prism and B is a pyramid. A prism with a
8 cm
base of sides has 3 edges. A pyramid
(3 8) 2 8 8 112 cm
with a base of sides has 2 edges. A
common multiple of 3 and 2 is 6. Draw a table
with the number of edges as multiples of 6.
Type 8-2 Dual Polyhedron p.72~p.73 Edges 6 12 18 24
1
Sides of the impossible 4 6 8
base of prism
Vertices of impossible 8 8 16
prism
Sides of the 3 6 9 12
base of pyramid
2 Regular octahedron Vertices of 4 7 10 13
pyramid
Difference in 123
vertices
3 Faces Vertices Edges
Cube 6 8 12 Therefore, A is a rectangular prism because it
8 6 12 has a rectangular base, and B is a hexagonal
pyramid because it has a hexagonal base.
4 Regular octahedron, vertices 3 At least three faces must meet at one vertex to
Problem solving make a solid figure. Remember that the angle of
an equilateral triangle is 60 . So if three, four, or
1 The dual of a regular tetrahedron is a regular five equilateral triangles meet at one vertex,
each interior angle becomes 180 , 240 , or
tetrahedron. 300 , respectively. It is impossible for more than
six faces to meet at a vertex because each
2 The dual of a regular octahedron is a cube. interior angle would be greater than 360 , which
is impossible. Therefore, only three regular
There are 6 faces and 8 vertices in a cube. polyhedrons: the tetrahedron, octahedron, and
Therefore, the sum is 6 8 14. icosahedron can be made.
Answer Key
