Area of a rectangle is 99 ft^2 , and the length of the rectangle is 7 ft more than twice the width, how do you find the dimensions of the rectangle?

1 Answer
Feb 9, 2016

Answer:

Width #=11/2# ft and Length #=18# ft

Explanation:

Let width of the rectangle be #x# ft
Given that length is #7# ft more than the twice the width.
#:.# length #=2xx "width"+7# ft
or Length #=2x+7# ft

Area of the rectangle = Length #xx# Width
Inserting given and above values

#99=(2x+7)xxx#
#=>2x^2+7x-99=0#
Solve by finding roots #x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-7+-sqrt(7^2-4*2*(-99)))/(2*2)#
or #x=(-7+-sqrt(49+792))/4#
or #x=(-7+-sqrt(841))/4#
or #x=(-7+-29)/4#
or #x=-9, 11/2#
Ignoring #-ve # root as width can not be negative.
Width #=11/2 and# Length #2xx11/2+7=18#