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Answer:

  (a)  58 m²

Step-by-step explanation:

The area of a composite figure is found by dividing it into simpler shapes whose area formulas are known.

Extending the horizontal line to the left, we find it divides the shape into an upper rectangle with dimensions 2 m wide and 4 m high, and a lower rectangle with dimensions 10 m wide and 5 m high.

The area of a rectangle is the product of its dimensions:

  A = LW . . . . . where L is the length and W is the width

The upper rectangle has an area of ...

  A = (4 m)(2 m) = 8 m²

The lower rectangle has an area of ...

  A = (10 m)(5 m) = 50 m²

The area of the figure is the sum of the areas of the upper and lower rectangles:

  A = (8 m²) +(50 m²) = 58 m²

The area of the figure is 58 square meters.

Additional comment

The area can also be figured by considering an enclosing rectangle 9 m high by 10 m wide, for an area of (9 m)(10 m) = 90 m². From this rectangle, a rectangle 4 m high and 8 m wide is removed from the upper right corner. The removed area is (4 m)(8 m) = 32 m². Then the remaining area is ...

  90 m² -32 m² = 58 m² . . . . the area of the figure shown

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Saved
(06.01 LC)
The value of 33 + 42 = ___. (5 points)

Question 1 options:
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Question 2 (5 points)
Saved
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What is the value of the expression shown below?

7 + (2 + 6) 2 ÷ 4 ⋅ (5 points)

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1)

23


2)

18


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9


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8

Question 3 (5 points)
Saved
(06.02 LC)
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Question 3 options:

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8y

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8 − y

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Question 4 (5 points)
Saved
(06.02 LC)
What best describes the expression ? (5 points)

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3 more than some number

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Some number times 3

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Some number divided by 3

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3 less than some number
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(06.03 LC)
What is the constant term in the expression 8xy + 9x2 y − 5x − 2? (5 points)

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(06.03 MC)
Which of the following best describes the expression 5(x + 2)? (5 points)

Question 6 options:

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The sum of constant factors 5 and x + 2

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The product of constant factors 5 and x + 2

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The sum of a constant factor 5 and a 2-term factor x + 2

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The product of a constant factor 5 and a 2-term factor x + 2
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2 was not multiplied by x

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Incorrect sign was used for the first term

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(06.07 LC)
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12p

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36p

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(06.07 MC)
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1)

A and B

2)

B and C

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A and D

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A and C

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