AMC 8 Daily Practice - Pythagorean Theorem

Complete problem set with solutions and individual problem pages

Problem 8 Easy

Quadrilateral ABCD is a rectangle. What is the area of the quadrilateral EGFH?

  • A.

    15

  • B.

    16

  • C.

    18

  • D.

    21

  • E.

    24

Answer:C

Given AB = 6 and EC = 8, in \triangle ABF, we find BF = 10.

The area of parallelogram BEDF can be expressed in two ways:

BE \times AB and BF \times EG.

Thus: 5 \times 6 = 10 \times EG

Solving for EG, we get EG = 3.

In the right triangle BGE, by the Pythagorean theorem: BG = \sqrt{BE^2 - EG^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4.

Since GF = 10 - 4 = 6 , the area of the quadrilateral EGFH is: GF \times EG = 6 \times 3 = 18.

Answer: The area of the quadrilateral EGFH is 18 square units.

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