Do you length of a rectangle is five times the width. If the length and width are both increased by 2, the new area would be 85. What is the original area?
Accepted Solution
A:
Answer:The original of the given rectangle = 45 sq units.Step-by-step explanation:Here, let us assume the actual width of the rectangle = aSo, the actual length of the rectangle = 5 x ( width) = 5 (a) = 5 aNow, the new width w' = ( a + 2)and the new length l' = ( 5 a + 2)AREA OF THE RECTANGLE = LENGTH x WIDTHSo, the area of the new rectangle = (a + 2)(5 a +2)Also, new area = 85 ⇒(a + 2)(5 a +2) = 85[tex]\implies 5a^2 + 10 a + 2a + 4 = 85\\\implies 5a^2 + 12a - 81 = 0\\\implies 5a^2 - 15 a + 27 a - 81 = 0\\\implies 5a(a -3) + 27(a -3) = 0\\\implies (5a +27)(a-3)= 0[/tex]⇒ ( 5a +27) =0 or (a-3) = 0⇒ a = -27/5 or a = 3But, a = Width of a rectangle , so a CANNOT be Negative⇒ a ≠ -27/5 and a = 3So, the actual width of the rectangle = a = 3The length of the rectangle = 5 a = 5 (3) = 15The original area = Original L x Original W = 3 x 15 = 45 sq units.