SOLUTION: Mr dean and Mr Tomlin have similar rectangular yards. the area of Mr Tomlin's yard is four times bigger than the area of Mr dean's yard. the length of Mr dean's yard is 15m and the

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Area of Rectangle of Mr dean's yard = Length x Width = Ld x Wd
Area of Rectangle of Mr Tomlin's yard = Length x Width = Lt x Wt
Because both rectangles are similar that means their lenghts and widths would be propotional.


Because Area of Rectangle of Mr Tomlin's yard = 4 x Area of Rectangle of Mr dean's yard
Lt x Wt = 4 (Ld x Wd)
Then
Lt x Wt = 2Ld x 2Wd
This means Length of Mr Tomlin's yard 2 times of Length of Mr Dean's yard and
width of Mr Tomlin's yard 2 times of width of Mr Dean's yard.
the length of Mr dean's yard is 15m
Then the length of Mr Tomlin's yard is 15x2 = 30 m
the area of Mr Tomlin's yard is 720m square = Lt x Wt = 30 x Wt
Wt = 720/30 = 24 m (the width of Mr Tomlin's yard)
Lt = 30 , Wt = 24
the lenght of fencing of Mr Tomlin's would be 2*(30+24)
and the cost of fencing of Mr Tomlin's would be 2*(30+24)*15 = $1620
For Mr Dean's yard
Ld = 15 m
Wd = half of the width of Mr Tomlin's = (1/2)*24 = 12 m
the lenght of fencing of Mr Dean's would be 2*(15+12)
and the cost of fencing of Mr Dean's would be 2*(15+12)*15 = $810