To solve this problem, we first need to define the dimensions of the two rectangles based on the information provided.
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Dimensions of the dark rectangle (on the left):
- Height = 7 units (as stated)
- Width = 2 units (as stated)
Thus, the area \( A_d \) of the dark rectangle is: \[ A_d = \text{Height} \times \text{Width} = 7 \times 2 = 14 \text{ square units} \]
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Dimensions of the light rectangle (on the right):
- Height = 7 units (same height as the dark rectangle)
- Width = \( x \) units (variable width given in the problem)
Thus, the area \( A_l \) of the light rectangle is: \[ A_l = \text{Height} \times \text{Width} = 7 \times x = 7x \text{ square units} \]
(a) Find the area of the following (in square units).
- Area of the dark rectangle (on the left): \[ A_d = 14 \text{ square units} \]
- Area of the light rectangle (on the right): \[ A_l = 7x \text{ square units} \]
(b) Give the area of the entire figure (in square units) in two different ways.
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As a sum of two areas: The total area \( A_t \) of the entire figure is the sum of the areas of the two rectangles: \[ A_t = A_d + A_l = 14 + 7x \text{ square units} \]
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As a product of the length and width: The entire figure is a rectangle with:
- Height = 7 units (common height for both rectangles)
- Width = \( 2 + x \) (the width of the dark rectangle plus the width of the light rectangle)
Thus, the area can also be calculated as: \[ A_t = \text{Height} \times \text{Width} = 7 \times (2 + x) = 14 + 7x \text{ square units} \]
In summary, the total area of the entire figure from both methods is given by:
- As a sum: \( A_t = 14 + 7x \)
- As a product: \( A_t = 14 + 7x \)
Both approaches yield the same total area.
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