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The area of a rectangle is 198 square ft. We know that one side is 15 ft longer than three times the other side. The rectangle has dimensions a and b where a is less than or equal to b. So a (the SHORTER side) is how many feet?

User Gelupa
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Given that the rectangle has dimensions a and b where a is less than or equal to b

Let b represent the length of the rectangle.

Let a represent the width of the rectangle.

Given that the area of the rectangle is 198 square feet, it means that

ba = 198 equation 1

We know that one side is 15 ft longer than three times the other side. It means that

b = 15 + 3a

Substituting b = 15 + 3a into equation 1, it becomes

a(15 + 3a) = 198

15a + 3a^2 = 198

3a^2 + 15a - 198 = 0

Dividing through the equation by 3, it becomes

a^2 + 5a - 66 = 0

We would solve the quadratic equation by applyling the method of factorisation. We would find two numbers such that their sum or difference is 5a and their product is - 66a^2. The numbers are 11a and 6a. The equation becomes

a^2 + 11a - 6a - 66 = 0

a(a + 11) - 6(a + 11) = 0

(a - 6)(a + 11) = 0

a - 6 = 0 or a + 11 = 0

a = 6 or a = - 11

Since the dimension of the rectangel cannot be negative, then the value of a is 6

Substituting a = 6 into equation 1, it becomes

6b = 198

b = 198/6

b = 33

The shorter side is 6 feet

The longer side is 33 feet

User Dogu Arslan
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