Answer:
1.3 m
Step-by-step explanation:
Given dimensions of rectangle ABCD:
- width = x
- length = (x + 4)
[tex]\begin{aligned}\textsf{Area of a rectangle} & = \sf width \times length\\\\\implies \sf Area\:of\:ABCD& = x(x+4)\\& = x^2+4x \end{aligned}[/tex]
Given dimensions of isosceles triangle ADE:
- Base = DE = x
- Height = AD = x
[tex]\begin{aligned}\textsf{Area of a triangle} & = \dfrac{1}{2} \times \sf base \times height\\\\\implies \textsf{Area of $\triangle$ADE}& = \dfrac{1}{2} \cdot x \cdot x\\& = \dfrac{1}{2}x^2\end{aligned}[/tex]
As the area of the portion remaining when ADE is cut off from ABCE is 6 m²:
[tex]\begin{aligned}\sf ABCD-ADE & = 6\\\implies x^2+4x-\dfrac{1}{2}x^2 & = 6\\2x^2+8x-x^2 & = 12\\x^2+8x & = 12\\x^2+8x-12 & = 0\end{aligned}[/tex]
Solve the found quadratic equation by completing the square.
Move the constant to the right side of the equation:
[tex]\implies x^2+8x-12+12=0+12[/tex]
[tex]\implies x^2+8x=12[/tex]
Add the square of half the coefficient of x to both sides:
[tex]\implies x^2+8x+\left(\dfrac{8}{2}\right)^2=12+\left(\dfrac{8}{2}\right)^2[/tex]
Simplify:
[tex]\implies x^2+8x+4^2=12+4^2[/tex]
[tex]\implies x^2+8x+16=12+16[/tex]
[tex]\implies x^2+8x+16=28[/tex]
Factor the perfect square trinomial on the left side:
[tex]\implies (x+4)^2=28[/tex]
Square root both sides:
[tex]\implies \sqrt{(x+4)^2}=\sqrt{28}[/tex]
[tex]\implies x+4=\pm \sqrt{28}[/tex]
[tex]\implies x+4=\pm \sqrt{4 \cdot 7}[/tex]
[tex]\implies x+4=\pm \sqrt{4}\sqrt{7}[/tex]
[tex]\implies x+4=\pm 2\sqrt{7}[/tex]
Subtract 4 from both sides:
[tex]\implies x=-4\pm 2\sqrt{7}[/tex]
As length is positive:
[tex]\implies x=-4+ 2\sqrt{7}\:\:\sf only[/tex]
Taking the value of √7 = 2.65:
[tex]\implies x \approx -4+ 2(2.65)[/tex]
[tex]\implies x \approx -4+5.3[/tex]
[tex]\implies x \approx 1.3[/tex]
Therefore, the width of the sheet is 1.3 m.
Learn more about completing the square here:
https://brainly.com/question/27933930
