AMC 8 Daily Practice Round 7

Complete problem set with solutions and individual problem pages

Problem 10 Medium

In parallelogram ABCD, E and F are the midpoints of AD and CD respectively. Given that EF = 4 and FB = 6, what is the area of the parallelogram?

  • A.

    20

  • B.

    25

  • C.

    32

  • D.

    40

  • E.

    64

Answer:C

Draw EG through point E parallel to CD, and draw FH through point F parallel to AD.

We observe that:

The area of \triangle ABE is \frac{1}{4} of the area of the parallelogram.

The area of \triangle BCF is \frac{1}{4} of the area of the parallelogram.

The area of \triangle DEF is \frac{1}{8} of the area of the parallelogram.

Therefore, the area of \triangle EFB is \frac{3}{8} of the area of the parallelogram.

The area of \triangle EFB is calculated as: \frac{4 \times 6}{2} = 12.

Let the area of the parallelogram be A.

Then: \frac{3}{8}A = 12.

Solving for A: A = 12 \div \frac{3}{8} = 32.

Thus, the area of the parallelogram is 32.

Related Problems
Problem 8 AMC 8 Daily Practice Round 7
Problem 9 AMC 8 Daily Practice Round 7
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Problem 12 AMC 8 Daily Practice Round 7
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