SOLUTION: Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm^2. Find the dime

Question 170636This question is from textbook Algebra structure and method book one
: Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm^2. Find the dimensions of the new rectangle. This question is from textbook Algebra structure and method book one

Found 2 solutions by checkley77, stanbon:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
20*23=460cm^2 is the original area.
(20-x)(23-x)=460-120
460-43x+x^2=340
x^2-43x+460-340=0
x^2-43x+120=0
(x-40)(x-3)=0
x-3=0
x=3
Thus the new dimentions are:
20-3=17
23-3=20
Proof:
17*20=460-120
340=340
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm^2. Find the dimensions of the new rectangle.
---------------------------
Original DATA:
area = 20*23 = 460 cm^2
--------------------------
New DATA:
area = (20-x)(23-x) cm^2
------------------------------
EQUATION:
original area - new area = 120 cm^2
460 -[460-43x+x^2] = 120
460 - 460 +43x - x^2 = 120
x^2 - 43x + 120 = 0
x = [43 +- sqrt(43^2 -4*120)}/2
x = [43 +- sqrt(1369)]/2
x = [43 +- 37]/2
Positive solution:
x = 3 cm or x = 40 cm
Only x = 3 cm is realistic.
=============================
Cheers,
Stan H.
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