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

"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"

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