Question 198910
sol:
the original dimensions are 15cm and 20cm
let the dimensions are decreased by x amount
after decreasing the dimensions are 15-x,20-x
area of the original rectangle=15*20=300cm^2
now the area of the new recatngle=(15-x)*(20-x)
according to the problem
area of the new rectangle=half the area of the original rectangle
 area of the new rectangle=1/2*area of the original rectangle
(15-x)(20-x)=1/2*300
(15-x)(20-x)=300/2
(15-x)(20-x)=150
15*(20-x)-x*(20-x)=150
[15*20-15*x]-[x*20-x*x]=150
[300-15x]-[20x-x^2]=150
300-15x-20x+x^2=150
300-35x+x^2=150
x^2-35x+300=150
x^2-35x+300-150=0
x^2-35x+150=0
x^2-5x-30x+150=0
x(x-5)-30(x-5)=0
(x-5)(x-30)=0
x-5=0
x=5
x-30=0
x=30
now we take x=5
the dimensions of the new rectangle=15-x,20-x
=15-5,20-5
=10,15
now we take x=30
the dimensions of the new rectangle=15-x,20-x
=15-30,20-30
=-15,-10
in this problems we always take positive values
 so the dimensions of the new rectangle were 10cm,15cm