Respuesta :
Answer:
The length of the rectangle is 11 inch and its width is 8 inch.
Step-by-step explanation:
Given:
Rectangle [tex](R_1)[/tex] and rectangle [tex](R_2)[/tex].
Let the width of the rectangle be 'x' inch.
Its length = '(x+3)' inch
Area of the rectangle = [tex](A_1)[/tex]
⇒[tex]A_1[/tex] = [tex](x)(x+3)[/tex]
Now for rectangle [tex](R_2)[/tex] its width is decreased by 1 inch and length is doubled,the area [tex](A_2)[/tex] will increased by 66 sq-inch.
So,
[tex]A_2=A_1+66[/tex]
According to the question.
⇒ Area of rectangle [tex](R_2)[/tex] with width '(x-1)' and length '2(x+3)'='(2x+6)' .
⇒ [tex]A_2=(x-1)(2x+6)[/tex]
Plugging the values of [tex]A_2[/tex] in terms of [tex]A_1[/tex].
⇒ [tex]A_1+66= (x-1)(2x+6)[/tex]
Plugging [tex]A_1[/tex] = [tex](x)(x+3)[/tex] into the equation.
⇒ [tex](x)(x+3)+66=(x-1)(2x+6)[/tex]
⇒ [tex]x^2+3x+66=2x^2+6x-2x-6[/tex]
⇒ [tex]x^2-2x^2+3x-6x+2x+6+66=0[/tex]
⇒ [tex]-x^2-x+72=0[/tex]
Now using quadratic formula we can find the value of 'x'.
Quadratic formula, (x)= [tex]\frac{\pm b+\sqrt{b^2-4ac} }{2a}[/tex]
x= -9 and x= 8
Discarding the negative value.
Our width = 'x' = 8 inches.
So the length of the rectangle = 11 inches and width of the rectangle = 8 inches.
