Page 169 - NUMINO Challenge_C1
P. 169
Answer
Key
4 The rectangle with the greatest area is a square Problem solving
with a side of 11 cm.
1 To make a quadrilateral with
Therefore, the area is 11 11 121 cm2.
the greatest area, the two
Problem solving 7 cm
diagonal lines should be 10 cm
1 The area of a rectangle is greatest when the
perpendicular, and the 6 cm
difference between the width and the length is 9 cm
the smallest. The sum of the lengths of the eight
sticks is (4 2) (5 4) (6 2) 40 cm, so the difference between the
area is greatest when one side is 40 4 10 cm.
Therefore, the greatest area of the rectangle is lengths of the diagonals should be the smallest.
10 10 100 cm2.
16 16 2 128 cm2
2 The perimeter of a rectangle with a fixed area
2
increases as the difference between the width
and the length increases. 25 cm
(Area) (Width) (Length), so the possible pairs
of (widths, lengths) are (40,1), (20,2), (10,4), 25 cm 25 cm
(8,5). The one with the greatest difference has
the greatest perimeter, so it should have a width 25 cm
of 40 cm and a length of 1 cm.
50 50 2 1250 cm2
Creative Thinking p.134~p.135
Type 15-2 Diagonals and Area p.132~p.133 1 When the perimeter is the smallest, the number
1 When the base is fixed, the triangle with the of sticks used is the smallest as well. The
greatest height has the greatest area. perimeter of a rectangle with a fixed area of
90 36 m2 is the smallest when the difference
between the two sides is the smallest.
2 (2 5, 9 12), (2 12, 5 9), (2 9, 5 12) 36 1 36 2 18 3 12 4 9 6 6,
(7cm, 21cm), (14cm, 14cm), (11cm, 17cm) so the difference is the smallest when the two
sides are each 6 m long. Therefore, you will
3 14 14 1 98 cm2 need 6 4 24 sticks.
2
2 The sum of the lengths of all the sticks is
(8 2) (5 6) (4 2) 54 m. Since all the
sticks are used, the sum of the width and the
length is 27 m. To make a rectangle with the
greatest area, the difference between the width
and the length must be the smallest. The width
must be longer than the length, so the width is
14 m and the length is 13 m.
Width: 5 5 4 14 m
Length: 8 5 13 m
Therefore, two 5 m sticks and one 4 m stick are
used to make the width.
NUMINO Challenge C1
Key
4 The rectangle with the greatest area is a square Problem solving
with a side of 11 cm.
1 To make a quadrilateral with
Therefore, the area is 11 11 121 cm2.
the greatest area, the two
Problem solving 7 cm
diagonal lines should be 10 cm
1 The area of a rectangle is greatest when the
perpendicular, and the 6 cm
difference between the width and the length is 9 cm
the smallest. The sum of the lengths of the eight
sticks is (4 2) (5 4) (6 2) 40 cm, so the difference between the
area is greatest when one side is 40 4 10 cm.
Therefore, the greatest area of the rectangle is lengths of the diagonals should be the smallest.
10 10 100 cm2.
16 16 2 128 cm2
2 The perimeter of a rectangle with a fixed area
2
increases as the difference between the width
and the length increases. 25 cm
(Area) (Width) (Length), so the possible pairs
of (widths, lengths) are (40,1), (20,2), (10,4), 25 cm 25 cm
(8,5). The one with the greatest difference has
the greatest perimeter, so it should have a width 25 cm
of 40 cm and a length of 1 cm.
50 50 2 1250 cm2
Creative Thinking p.134~p.135
Type 15-2 Diagonals and Area p.132~p.133 1 When the perimeter is the smallest, the number
1 When the base is fixed, the triangle with the of sticks used is the smallest as well. The
greatest height has the greatest area. perimeter of a rectangle with a fixed area of
90 36 m2 is the smallest when the difference
between the two sides is the smallest.
2 (2 5, 9 12), (2 12, 5 9), (2 9, 5 12) 36 1 36 2 18 3 12 4 9 6 6,
(7cm, 21cm), (14cm, 14cm), (11cm, 17cm) so the difference is the smallest when the two
sides are each 6 m long. Therefore, you will
3 14 14 1 98 cm2 need 6 4 24 sticks.
2
2 The sum of the lengths of all the sticks is
(8 2) (5 6) (4 2) 54 m. Since all the
sticks are used, the sum of the width and the
length is 27 m. To make a rectangle with the
greatest area, the difference between the width
and the length must be the smallest. The width
must be longer than the length, so the width is
14 m and the length is 13 m.
Width: 5 5 4 14 m
Length: 8 5 13 m
Therefore, two 5 m sticks and one 4 m stick are
used to make the width.
NUMINO Challenge C1
