AMC 8 Daily Practice - Area Tricks

Complete problem set with solutions and individual problem pages

Problem 1 Easy

The area of quadrilateral ABCD is 25. It is known that AD = DC, DE = BE, and AE = 2. What is the area of the shaded part?

  • A.

    15

  • B.

    18

  • C.

    19

  • D.

    20

  • E.

    22

Answer:D

 

Since AE = CD, as shown in the figure, draw a line DF through point D perpendicular to the extension of BC at point F.

It can be obtained that \triangle ADE \cong \triangle CDF (congruent triangles).

Therefore, quadrilateral ABCD can be transformed into square DFBE.

The area of square DFBE is 25, so DE = EB = 5.

Then, BC = BF - CF = BF - AE = 5 - 2 = 3.

Thus, the area of the shaded part is: \frac{5 \times (5 + 3)}{2} = 20

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Problem 2 Easy

There is a rectangular flower bed. In the middle of the flower bed, there is a pedestrian path with a width of 3 meters. The flower bed is 53 meters long and 25 meters wide. What is the area of the pedestrian path in square meters?

  • A.

    200

  • B.

    225

  • C.

    234

  • D.

    245

  • E.

    260

Answer:B

Through observation, we can translate the path to the position as shown in the following figure.

Therefore, the area of the pedestrian path is calculated as: 25 \times 3 + (53 - 3) \times 3 = 225

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Problem 3 Easy

As shown in the figure below, the side length of square ABCD is 10 centimeters, and the rectangle EFGH has a length of 8 centimeters and a width of 5 centimeters. What is the difference between the areas of the two shaded parts?

  • A.

    15

  • B.

    20

  • C.

    30

  • D.

    40

  • E.

    60

Answer:E

 

As shown in the figure, let the areas of the three regions be denoted as , , and respectively.

We have:

Ⅰ = \text{Area of square } ABCD - Ⅱ

Ⅲ = \text{Area of rectangle } EFGH - Ⅱ

Then, the difference between the areas of the two shaded parts is: Ⅰ - Ⅲ = \left( \text{Area of square } ABCD - Ⅱ \right) - \left( \text{Area of rectangle } EFGH - Ⅱ \right) = \text{Area of square } ABCD - \text{Area of rectangle } EFGH = 10 \times 10 - 8 \times 5 = 100 - 40 = 60

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Problem 4 Easy

Rectangle ABCD has a length of 8 centimeters and a width of 4 centimeters. Its diagonals AC and BD divide it into four parts, and diagonal AC is further divided into four equal segments. What is the area of the shaded part?

  • A.

    6cm^{2}

  • B.

    7cm^{2}

  • C.

    8cm^{2}

  • D.

    9cm^{2}

  • E.

    10cm^{2}

Answer:C

According to the formula for the area of a triangle: \text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}.

The area of triangle = the area of triangle = the area of triangle = the area of triangle .

The area of the shaded part is calculated as: \left( \frac{4 \times 4} {2} \right)\div 2 \times 2 = 8cm^{2}

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Problem 5 Medium

Square ABCD has a side length of 6 centimeters. It is known that BE = BF, EG = 3 centimeters, and FG = 5 centimeters. What is the area of the shaded part?

  • A.

    20

  • B.

    21

  • C.

    24

  • D.

    25

  • E.

    26

Answer:A

Since BE = BF, we can fold the figure as shown, resulting in a new figure.

Through this folding, the area of the shaded part can be calculated as follows: \frac{144 - 64}{4} = 20.

Thus, the area of the shaded part is 20 square centimeters.

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Problem 6 Easy

The area of a regular octagon is 40 square centimeters. What fraction of the area of the octagon is shaded?

  • A.

    \frac{1}{3}

  • B.

    \frac{15}{32}

  • C.

    \frac{1}{2}

  • D.

    \frac{3}{5}

  • E.

    \frac{9}{16}

Answer:C

By means of division, the octagon is divided into 4 rectangles and 8 triangles.

The shaded part contains 2 rectangles and 4 triangles.

Therefore, the area of the shaded part accounts for \frac{1}{2} of the total area.

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Problem 7 Easy

A large square and a small square are placed side by side, with the side length of the small square being 6cm. What is the area of the shaded part?

  • A.

    25 cm^{2}

  • B.

    28 cm^{2}

  • C.

    32 cm^{2}

  • D.

    36 cm^{2}

  • E.

    40 cm^{2}

Answer:D

As shown in the figure, connect the diagonals of the two squares.

Since the two diagonals are parallel, we find that the area of triangle is equal to the area of triangle .

Therefore, the area of the shaded part is converted to the area of the small square, which is: 6 \times 6 = 36 cm^{2}

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Problem 8 Medium

In rectangle 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 rectangle?

  • 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 rectangle.

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

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

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

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

Let the area of the rectangle be A.

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

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

Thus, the area of the rectangle is 32.

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