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VARVARA [1.3K]
3 years ago
8

The length of a rectangle is 3 feet more than twice the width. The perimeter is 128 feet. Find the length and width.

Mathematics
1 answer:
iragen [17]3 years ago
7 0
We are going to consider a rectangle with length called "a" and width "b"  Then we know that perimeter is the sum of both sides 2*a + 2*b = 128  And also the following equation given by the question: a = 2*b + 3  So solving the first equation: 2*(2*b+3) + 2*b = 128 4*b+6+2*b=128 6*b = 122 b = 122/6 = 61/3  Then a=2*122/6+3 = 244/6+3 = 262/6 = 131/3  Solution:  Length = 131/3 Width = 61/3
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In case you’ve forgotten, the Pythagorean Theorem states that, in any given right triangle, the sum of the squares of the lengths of its legs is equal to the square of the length of its hypotenuse (the side opposite its right angle). If we call the lengths of the legs a and b, and the length of the hypotenuse c, this can be expressed in notation as a^2+b^2=c^2 (it doesn’t matter in this case which leg you pick for a and which you pick for b). Here, if we choose the left leg as a and the bottom leg as b, we’re given that a^2 (the area of a square with sides of length a) is 25 sq. in, and b is 3.5 in. Plugging those values into the equation, we have:

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From here, you don’t even need to solve for c, you just need to find the value of c^2 (since you’re trying to find the area of a square with side lengths c). Just solve the left side of the equation, and you’ll have your answer in square inches.
























7 0
3 years ago
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The 8 goes to the root and the 3x becomes the exponent

4 0
3 years ago
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