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Perimeter: Problems with Solutions
By Catalin David
Problem 1
What is the perimeter of triangle ABC?
Solution:
The perimeter is the sum of the lengths of all sides
P = 28 + 51 + 46 = 79 + 46 = 125 cm
Problem 2
The red triangle is equilateral with a side of 23 centimetres.
Its perimeter is
cm.
Solution:
All three sides of an equilateral triangle are equal. Thus, its perimeter is 23 × 3 = 69 см.
Problem 3
An isosceles triangle has a perimetеr of 37 centimetres and its base has a length of 9 centimetres. Each of the other two sides has a length of
cm.
Solution:
The isosceles triangle has two equal sides. The sum of the equal sides is 37 - 9 = 28 cm. Thus, one of them is 28 ÷ 2 = 14 cm
Problem 4
Tim's garden is shaped like a square whose side is 9 meters. What's the length of the fence which surrounds the garden?
Answer:
meters.
Solution:
All sides of the square are equal. The length of the fence is 4 times the side.
P = 4 × 9 = 36 meters
Problem 5
In rectangle ABCD the red side is 18 cm and the blue side is 12 cm. The perimeter is
cm.
Solution:
Opposite sides of the rectangle are equal.
The perimeter is 2 × 18 + 2 × 12 = 36 + 24 = 60 cm.
Problem 6
A rectangle has a length of 28 cm and a width smaller by 6 cm. Its perimeter is
cm.
Solution:
The width is 28 - 6 = 22 cm
The perimeter is 2 × (28 + 22) = 2 × 50 = 100 cm
Problem 7
A rectangle has a width of 19 cm and a length three times greater than its width. Its perimeter is
cm.
Solution:
The length is 19 × 3 = 57 cm.
The perimeter is 2 × (57 + 19) = 2 × 76 = 152 cm.
Problem 8
The width and length of a rectangle are two consecutive numbers. If the perimeter is 30 cm, then its width is
cm.
Solution:
The perimeter is 2 × (L + A) = 30
L + W = 30 ÷ 2 = 15
7 + 8 = 15
W = 7
Problem 9
John builds a square with a side of 12 cm from some wire. If from the same wire, he builds a rectangle with a width of 6 cm, what is the length of the rectangle?
Solution:
If the side of the square is 12 cm, then its perimetеr will be 48 cm. We build the rectangle from the same wire, so it is the same perimetеr. Its width is 6 cm, so its length is (48-12) ÷ 2 = 36 ÷ 2 = 18 cm
Problem 10
A park shaped like a rectangle has a length of 24 meters and a width of 18 meters. If trees are planted on its sides with gaps of 2 meters between them, how many trees are needed?
Solution:
The perimeter of the rectangle is 2 × (24 + 18) = 2 × 42 = 84 meters, so there will be 84 ÷ 2 = 42 trees.
Problem 11
What is the perimeter of the blue figure?
Solution:
There are 2 squares which share a part of a side. Because the side of the square is 10 cm and a part of it is 8 cm, then the common part is 2 cm and the leftover part from the side of the other square is 8 cm too.
Perimeter is 10 + 10 + 8 + 10 + 10 + 10 + 8 + 10 = 76 cm.
Problem 12
What is the perimeter of the red figure?
Solution:
Because the white figure is a rectangle, the unknown sides of the red figure are 12 cm, 8cm, 6cm, 16 cm. Perimeter is 18 + 24 + (12 + 6) + (8 + 16) = 18 + 24 + 18 + 24 = 42 + 42 = 84 cm
Problem 13
Can we built a triangle in which a side is half the perimeter?
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.
Problem 14
The square in the figure has a perimeter of 24 cm. The blue triangle has a perimeter of 15 cm. What is the perimeter of the red figure?
Solution:
If the perimeter of the square is 24 cm, then its side will be 24 ÷ 4 = 6 cm. If the perimeter of the triangle is 15 cm and the length of one of its sides is 6 cm, then the total length of the other two is 15 - 6 = 9 cm. The perimeter of the red figure is 9 + 6 + 6 + 6 = 27 cm.
Problem 15
Rectangle ABCD has a perimeter of 24 cm and diagonal has a length of 8 cm. What is the perimeter of the green triangle?
Solution:
If the perimeter of the rectangle is 24 cm, 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 cm.
Problem 16
The red figures are two triangles, each having a perimeter of 47 cm. The blue figure is a square. What is the distance covered by an ant on the path A-B-C-D-E-F-A?
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 centimeters.
Problem 17
What is the perimeter of the figure displayed in the image?
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 cm.
Problem 18
What is the perimetеr of the green zone if the blue zone has a width of 3 metres?
Solution:
The length of the green zone is 38 - (3 + 3) = 38 - 6 = 32 m.
The width of the green zone is 24 - (3 + 3) = 24 - 6 = 18 m.
The perimetre is 2 × (32 + 18) = 2 × 50 = 100 m.
Problem 19
The blue square has a perimetеr of 16 cm. The red triangles are equilateral. What distance is covered by a snail following path ABCDFGHA?
Solution:
Since the perimeter of the square is 16 cm, its side is 16 ÷ 4 = 4 cm. The triangles are equilateral, so their sides are equal. Thus, the sides of the triangles are also 4 cm. The distance covered by the snail is 7 × 4 = 28 cm.
Problem 20
The green squares each have a perimetеr of 12 cm. What is the perimetеr of the big square?
Solution:
If the perimeter of the small squares is 12 cm, then the side of a square is 3 cm. The side of the big square is three times bigger, so it is 9 cm.
The perimeter is 9 × 4 = 36 cm
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