(a) The length of a vertangle in three Inches more than the width. The area of the rectangle is 270 inches. Pind the Width of the revinugle A) Min (5) 15

To find the width of the rectangle, we need to use the given information about the length and the area.<br /><br />Let's assume the length of the rectangle is L and the width is W. According to the problem, the length is 3 inches more than the width. So, we can write the equation as:<br /><br />L = W + 3<br /><br />The area of a rectangle is given by the formula: Area = Length x Width. We are given that the area is 270 square inches. Therefore, we can write the equation as:<br /><br />270 = L * W<br /><br />Now, we can substitute the value of L from the first equation into the second equation:<br /><br />270 = (W + 3) * W<br /><br />Expanding the equation, we get:<br /><br />270 = W^2 + 3W<br /><br />Rearranging the equation, we have:<br /><br />W^2 + 3W - 270 = 0<br /><br />This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it using the quadratic formula:<br /><br />W = (-b ± √(b^2 - 4ac)) / (2a case, a = 1, b = 3, and c = -270. Plugging these values into the quadratic formula, we get:<br /><br />W = (-3 ± √(3(1)(-270))) / (2(1))<br /><br />Simplifying further, we have:<br /><br />W = (-3 ± √(9 + 1080)) / 2<br /><br />W = (-3 ± √1089) / 2<br /><br />W = (-3 ± 33) / 2<br /><br />Now, we have two possible solutions for W:<br /><br />W = (-3 + 33) / 2 = 30 / 2 = 15<br /><br />W = (-3 - 33) / = -36 / 2 = -18<br /><br />Since the width cannot be negative, we discard the second solution. Therefore, the width of the rectangle is 15 inches.<br /><br />So, the correct answer is:<br /><br />B) 15