Chapter ISG, Problem 9.6.6P

To determine

To find area of rectangle.

Answer to Problem 9.6.6P

width is $24.14ft$

Explanation of Solution

Given information:

Length of rectangle is $w-20$

Width of rectangle is $w$

Area of rectangle is $A=100f{t}^2$

Formula used:

  $A=lw$

Where

Length of rectangle is l

Width of rectangle is $w$

Calculation:

Length of rectangle, $l=w-20$

Width of rectangle $=w$

Area of rectangle, $A=100f{t}^2$

Using formula

  $A=(w-20)(w)$

put value of area

  $100=(w-20)(w)$

On solving,

  $100=w^2-20w$

On solving,

  $w^2-20w-100=0$

On solving,

  $w=\frac{-(-20)\pm \sqrt{{(-20)}^2-4(1)(-100)}}{2(1)}$

On solving,

  $w=\frac{20\pm \sqrt{400+400}}{2}$

On solving,

  $w=\frac{20\pm \sqrt{800}}{2}$

On solving,

  $w=\frac{20\pm 20\sqrt{2}}{2}$

On solving,

  $w=\frac{20+20\sqrt{2}}{2}$

  $w=\frac{20-20\sqrt{2}}{2}$

On solving,

  $w=10+10\sqrt{2}$

  $w=10-10\sqrt{2}$

On solving,

  $w\approx 24.14ft$

  $w\approx -4.14ft$

Width cannot be negative

Hence, width is $24.14ft$