The length of a rectangle is 8 centimeters less than three times its width. Its area is 35 square...

Question:

The length of a rectangle is 8 centimeters less than three times its width. Its area is 35 square centimeters. Find the dimensions of the rectangle.

Rectangular Area:

A rectangle is a figure that has 4 sides. For example, the tennis court which at has two longer sides which are parallel and equal and two shorter sides which are parallel and equal. The rectangular area is, therefore, the area covered by the rectangular shape.

The area of a rectangle is given by:

• {eq}A = l\times w {/eq}

Where {eq}l {/eq} is the length and {eq}w {/eq} is the width.

Let the width of the rectangle described in our question be:

• {eq}w = x\, \rm cm {/eq}

If the length is 8 cm less than 3 times the width, then we have the expression of the length as:

• {eq}l = (3x - 8)\, \rm cm {/eq}

The area of this rectangle is therefore equal to:

• {eq}A = x(3x - 8) = 3x^2 - 8x {/eq}

If the area is {eq}A = 35\, \rm cm^2 {/eq}, then we have:

• {eq}35= 3x^2 - 8x {/eq}
• {eq}3x^2 - 8x - 35 = 0 {/eq}

We have a quadratic equation to solve.

To solve the equation by factorization, we need to numbers such that their sum is {eq}-8 {/eq} and their product is {eq}3\times -35 = -105 {/eq}. These two numbers are {eq}-15 {/eq} and {eq}7 {/eq}.

Therefore:

• {eq}3x^2 + 7x - 15x - 35 = 0 {/eq}
• {eq}x(3x - 7) - 5(3x + 7)= 0 {/eq}
• {eq}(3x - 7)(x - 5) = 0 {/eq}
• {eq}x = 5, \quad x = -\dfrac{7}{3} {/eq}

We will consider the positive value of x since the length cannot be negative. Therefore:

• {eq}w = x = \boxed{5\, \rm cm} {/eq}
• {eq}l = 3(5) - 8 = \boxed{7\, \rm cm} {/eq}