AMC 8 Daily Practice Round 7

Complete problem set with solutions and individual problem pages

Problem 5 Easy

A figure is formed by arranging 22 small pieces of paper of the same size as shown in the diagram. Given that the length of each small piece of paper is 15 centimeters, what is the total perimeter of the shaded parts in the figure?

  • A.

    15cm

  • B.

    20cm

  • C.

    45cm

  • D.

    60cm

  • E.

    75cm

Answer:D

As shown in the diagram, we can observe that 5 lengths =3 lengths +3 widths.

Therefore, 2 lengths =3 widths.

Given that the length is 15\text{ cm}, the width can be calculated as follows: \text{Width} = 15 \times 2 \div 3 = 10cm.

The perimeter of one shaded part is \text{length} - \text{width} = 15 - 10 = 5 cm.

Thus, the total perimeter of the shaded parts is: 5 \times 4 \times 3 = 60cm.

Related Problems
Problem 3 AMC 8 Daily Practice Round 7
Problem 4 AMC 8 Daily Practice Round 7
Problem 6 AMC 8 Daily Practice Round 7
Problem 7 AMC 8 Daily Practice Round 7
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