The length of a rectangle is 5 yd less than double the width, and the area of the rectangle is 52 yd^2 . How do you find the dimensions of the rectangle?

2 Answers
Aug 18, 2016

Answer:

Width = 6.5 yds, length = 8 yds.

Explanation:

Define the variables first.

We could use two different variables, but we have been told how the length and width are related.

Let the width be #x" width is the smaller side"#
The length = #2x -5#

"Area = l x w" and the area is given to be 52 squ yards.

#A = x(2x-5) = 52#

#2x^2 -5x = 52" quadratic equation"#

#2x^2 -5x -52 = 0#

To factorise, find factors of 2 and 52 which cross-multiply and subtract to give 5.

#color(white)(xxx)(2)" "(52)#
#color(white)(xx.x) 2" 13 "rArr 1xx13 = 13#
#color(white)(xx.x) 1" 4 "rArr2xx4 = 8" "13-8 = 5#

We have the correct factors, now fill in the signs. We need -5.

#color(white)(xxx)(2)" "(-52)#
#color(white)(xx.x) 2" - 13 "rArr 1xx-13 = -13#
#color(white)(xx.x) 1" +4 "rArr2xx+4 = +8" "-13+8 = -5#

#(2x-13)(x+4) = 0#

Each factor could be equal to 0

#x = 6.5 or x = -4 # (reject)

The width = 6.5 yards. Now find the length: 6.5 x 2 -5 = 8 yards

Check:
Width = 6.5yds, length = 8yds
Area = 6.5 x 8 = 52

Aug 18, 2016

Answer:

Length#= 8 yd#
Width #= 6.5 yd#.

Explanation:

Let width be #=x#
Therefore, length #=2x -5#

We know that
#"Area" = "Length" xx "Width"#
Inserting given and assumed numbers we get

#52=(2x-5)xx x#
rearranging we obtain

#2x^2 -5x -52 = 0#

To factorize we use split the middle term method. We have two parts of middle term as #-13x and 8x#. The equation becomes

#2x^2-13x+8x-52 = 0#
Paring and taking out common factors we have
#x(2x-13)+4(2x-13) = 0#
#=>(2x-13)(x+4) = 0#

Setting each factor equal to #0#, we have two roots
#(2x-13)=0and (x+4) = 0#
#x = 13/2=6.5#
# x = -4 #, rejected as width can not be a #-ve# value

#:.#Width #= 6.5 yd#. And length#= 2xx6.5 -5 = 8 yd#

Check:
Area #= 8xx 6.5 = 52yd^2#