Question

PLEASEEEE ANSWER ASAP!!!!

Answer 1

Refer to the attached image.

Let the width of the border be 'x' inch.

Area of border = Area of rectangle ABCD - Area of rectangle PQRS

Area of rectangle ABCD = $length \times width$

= $AB \times BC$

= $(13+2x) \times (9+2x)$ square units.

Area of rectangle PQRS = $PQ \times RQ$

= $13 \times 9$

= 117 square units.

So, area of border = $(13+2x)(9+2x)-117 = 48$

As it is given that the area of border is 48 square units.

So, $117+26x+18x+4x^2-117 = 48$

$44x + 4x^2-48=0$

$4x^2 + 44x = 48$

So, the complete equation is $4x^2 + 44x = 48$.

Now, we will find 'x' in the equation $x^2+11x-12=0$,

$x^2 + 12x - x -12 =0$

$x(x+12)-1(x+12)=0$

(x-1) (x+12)=0

So, x = 1 or x = -12

Width can'not be negative.

So, x = 1 .

Therefore, the width of the border can'not be 2.5 inch.

Therefore, the width of the border is 1 inch.