Question 380474
The perimeter of a rectangle is 50m. If the width was doubled and the length was increased by 20m, the perimeter would be 100m. What are the length and width of the rectangle?
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let width be w
length = l
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Area A = l*w=50 m^2
l= 50/w
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second rectangle
width = 2w
length = l+20
Perimeter of second rectangle =2*(2w+(l+20)) 
P=2(2w+l+20)
P=4w+2l+40
plug the value of l above
P=4w+2(50/w)+40
100=4w+100/w+40
multiply by w
100w=4w^2+100+40w
re write
4w^2-100w+40w+100=0
4w^2-60w+100=0
/4
w^2-15w+25=0
find the roots w1,w2 using quadratic formula
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discriminant = 125
w1=((15+sqrt(125))/2)
w1=13.1 meters ... the width
w2 will be negative so ignore
length = 50/13.1
length = 3.82 meters
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CHECK perimeter of second rectangle
plug w & l 
2*(2*13.1+((3.82+20))=100
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m.ananth@hotmail.ca