Perimeter: Problems with Solutions

Solution:
The perimeter is the sum of the lengths of all sides
P = 28 + 51 + 46 = 79 + 46 = 125 in

Solution:
All three sides of an equilateral triangle are equal. Thus, its perimeter is 23 × 3 = 69 in.

Solution:
The isosceles triangle has two equal sides. The sum of the equal sides is 37 - 9 = 28 in. Thus, one of them is 28 ÷ 2 = 14 in

Solution:
All sides of the square are equal. The length of the fence is 4 times the side.
P = 4 × 9 = 36 feet

Solution:
Opposite sides of the rectangle are equal.
The perimeter is 2 × 18 + 2 × 12 = 36 + 24 = 60 in.

Solution:
The width is 28 - 6 = 22 in
The perimeter is 2 × (28 + 22) = 2 × 50 = 100 in

Solution:
The length is 19 × 3 = 57 in.
The perimeter is 2 × (57 + 19) = 2 × 76 = 152 in.

Solution:
The perimeter is 2 × (L + A) = 30
L + W = 30 ÷ 2 = 15
7 + 8 = 15
W = 7

Solution:
If the side of the square is 12 in, then its perimeter will be 48 in. We build the rectangle from the same wire, so it is the same perimeter. Its width is 6 in, so its length is (48-12) ÷ 2 = 36 ÷ 2 = 18 in

Solution:
The perimeter of the rectangle is 2 × (24 + 18) = 2 × 42 = 84 feet, so there will be 84 ÷ 2 = 42 trees.

Solution:
There are 2 squares which share a part of a side. Because the side of the square is 10 in and a part of it is 8 in, then the common part is 2 in and the leftover part from the side of the other square is 8 in too.
Perimeter is 10 + 10 + 8 + 10 + 10 + 10 + 8 + 10 = 76 in.

Solution:
Because the white figure is a rectangle, the unknown sides of the red figure are 12 in, 8 in, 6 in, 16 in. Perimeter is
18 + 24 + (12 + 8) + (6 + 16) = 18 + 24 + 20 + 22 = (22 + 18) + (24 + 20) = 40 + 44 = 84 in

Solution:
No, because the sum of two of the sides is equal to the third side. This case is possible only if the three points are situated on the same line.

Solution:
If the perimeter of the square is 24 in, then its side will be 24 ÷ 4 = 6 in. If the perimeter of the triangle is 15 in and the length of one of its sides is 6 in, then the total length of the other two is 15 - 6 = 9 in. The perimeter of the red figure is 9 + 6 + 6 + 6 = 27 in.

Solution:
If the perimeter of the rectangle is 24 in, then the sum of its length and its width is 24 ÷ 2 = 12. The perimeter of the green triangle is AC + CD + DA = 8 + 12 = 20 in.

Solution:
The covered distance is AB + BC + CD + DE + EF + FA. The perimeter of triangle ABF is AB + BF + FA = 47. The perimeter of triangle CDE is CD + DE + EC = 47. Because the blue figure is a square, BF = BC and CE = EF. Then AB + BC + FA = 47 and CD + DE + EF = 47. The covered distance is 47 + 47 = 94 inches.

Solution:
If we align the red segments we'll get a segment equal to the width of the rectangle. If we align the blue segments we'll get a segment equal to the length of the rectangle. The perimeter of the figure is equal to the perimeter of the rectangle = 2 × (27 + 19) = 2 × 46 = 92 in.

Solution:
The length of the green zone is 38 - (3 + 3) = 38 - 6 = 32 ft.
The width of the green zone is 24 - (3 + 3) = 24 - 6 = 18 ft.
The perimetre is 2 × (32 + 18) = 2 × 50 = 100 ft.

Solution:
Since the perimeter of the square is 16 in, its side is 16 ÷ 4 = 4 in. The triangles are equilateral, so their sides are equal. Thus, the sides of the triangles are also 4 in. The distance covered by the snail is 7 × 4 = 28 in.

Solution:
If the perimeter of the small squares is 12 in, then the side of a square is 3 in. The side of the big square is three times bigger, so it is 9 in.
The perimeter is 9 × 4 = 36 in