A square was altered so that one side is increased by 9 inches in the other side is decreased by 2 inches. The area of the resulting rectangle is 6o in 2 what was the area of the

area of the original square = 36 in²

Step-by-step explanation:

given data

one side is increased = 9 inches

other side is decreased = 2 inches

area of the resulting rectangle = 60 in²

solution

we consider here x as the length of any one side

and

longer side of the resulting rectangle = x + 9

and

the shorter side x - 2

so that area of this rectangle is (s+9) × (s-2) = 60 in²

standard form of equation is  x² + 7x - 18 = 60   1

it will be

x² + 7x - 78 = 0

(s+13) (s-6) = 0

so that here x  = -13

and  x = 6

here we Discard x = -13 because the side length cannot be negative.

so x = 6

and area of the original square = 36 in²

its d but looking at that hurts my eyes frfr

step-by-step explanation:

14

step-by-step explanation:

14 x 3 = 42