SOLUTION: David drew a rectangle is 143 square mm. Derek drew a rectangle inside of David which is 35 square mm. Derek's rectangle has a three mm border between his and David's rectangle. Wh

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: David drew a rectangle is 143 square mm. Derek drew a rectangle inside of David which is 35 square mm. Derek's rectangle has a three mm border between his and David's rectangle. Wh      Log On


   

Question 87249: David drew a rectangle is 143 square mm. Derek drew a rectangle inside of David which is 35 square mm. Derek's rectangle has a three mm border between his and David's rectangle. What are the length and width of David rectangle? -Jason
Answer by malakumar_kos@yahoo.com(315) About Me  (Show Source):
You can put this solution on YOUR website!
let the length &width of derek's rectangle be = l & b
area of the rectangle = 35 sq mm
length & width of David's rectangle = (l+3)&(b+3)
area of David's rectangle = 143 sq mm
(l+3)(b+3)=143 lb+3l+3b+9 =143
35+3(l+b)+9 = 143
3(l+b)= 143-44
3(l+b)=99
l+b = 99/3 = 33 l = 33-b--------eq 1
l.b = 35 b.(33-b)=35
33b-bsquare = 35
b(square)-33b+35=0
use x=(-b+-sqrt(b^2-4ac)/2.a b=33+-sqrt33^2-4.1.(-35)/2
b= 33+-sqrt1229/2
b= 33+-35.05/2
b = 68.05/2=34.03
or b=-2.05/2= -1.03
l=34.03 if b=-1.03 and l=-1.03 if b=34.03
length &width of David's rectangle = l+3=37.03 b=1.97 or vice versa