Using a system of equations from the equation of a rectangle's perimeter, it is found that:
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The perimeter of a rectangle of length l and width w is given by:
[tex]P = 2(l + w)[/tex]
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The perimeter of a rectangle is 141 feet
This means that:
[tex]2(l + w) = 141[/tex]
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Length is twice the width
This means that [tex]l = 2w[/tex]
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Finding the width:
Replacing into the first equation:
[tex]2(l + w) = 141[/tex]
[tex]2(2w + w) = 141[/tex]
[tex]2(3w) = 141[/tex]
[tex]6w = 141[/tex]
[tex]w = \frac{141}{6}[/tex]
[tex]w = 23.5[/tex]
The width of the rectangle is of 23.5 feet.
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The length is:
[tex]l = 2w = 2(23.5) = 47[/tex]
The length of the rectangle is of 47 feet.
A similar problem is given at https://brainly.com/question/16642085
Dimensions of the rectangle are:
length . b = 47 f
width . a = 23.5 f
Area of the rectangle A = 1104.5 ft²
Perimeter:
The Perimeter (p) of a rectangle of sides "a" and "b" is:
p = 2×a + 2×b (1)
If we affirm that the length is "b" it is twice "a" or
2×a = b (2)
Now we don't know "a" and "b" but we have two conditions, the parameters need to follow these two conditions and they are express mathematically in the equations (1) and (2)
p = 2×a + 2×b ⇒ . 141 = 2×a + 2×b . and . b = 2×a
We get a two equations system, solving by substitution of equation (2) in equation (1)
141 = 2×a + 2 × (2 ×a) . ⇒ 141 = 2×a + 4×a
6×a = 141
a = 141/6
a = 23.5 f. and b = 2×a b = 2× 23.5 . b = 47 f
Area of the rectangle is A = 23.5 × 47
A = 1104.5 ft²
