The length of a rectangle is 10 feet less than 3 times its width. How do you find the dimensions of this rectangle if the area is 48 square feet?

2 Answers
Feb 14, 2018

Answer:

#"length "=8" feet and width "=6" feet"#

Explanation:

#"let the width "=x#

#"then the length "=3x-10larrcolor(blue)"10 less than 3 times width"#

#• " area of rectangle "=" length "xx" width"#

#rArr"area "=x(3x-10)=3x^2-10x#

#"now area "=48#

#rArr3x^2-10x=48larrcolor(blue)"rearrange and equate to zero"#

#3x^2-10x-48=0#

#"the factors of - 144 which sum to - 10 are - 18 and + 8"#

#"splitting the middle term gives"#

#3x^2-18x+8x-48=0larrcolor(blue)"factor by grouping"#

#color(red)(3x)(x-6)color(red)(+8)(x-6)=0#

#(x-6)(color(red)(3x+8))=0#

#"equate each factor to zero and solve for x"#

#x-6=0rArrx=6#

#3x+8=0rArrx=-8/3#

#x>0rArrx=6#

#rArr"width "=x=6" feet"#

#rArr"length "=3x-10=18-10=8" feet"#

Answer:

Width#=6# feet and length #=8# feet

Explanation:

Let the width #= x# feet

So, length #= 3x-10# feet

Now area of rectangle

#= "length" xx "width"# sq unit.

#= (3x-10)x# sq feet.

Now as per question,

#(3x-10)x = 48#

#rArr 3x^2-10x-48=0#

#rArr 3x^2-(18-8)x-48=0#

#rArr 3x^2-18x+8x-48=0#

#rArr 3x(x-6)+8(x-6)=0#

#rArr (3x+8)(x-6)=0#

#rArr 3x+8=0, x-6=0#

#rArr x = -8/3 and 6.#

Width cannot be negative.

So, #x = 6#

Hence width is #6# feet, and length is #3 xx 6-10 = 8# feet