Respuesta :
here,
let width(b)be x then,
length (l)=9cm+x
area =112 sq cm
now,
area of rectangle=l*b
or, 112=(9+x)x
or, 112=9x+x^2
or, 0=x^2+9x-112
or, 0=x^2+(16-7)x-112
or, 0=x^2+16x-7x-112
or, 0=x(x+16)-7(x+16)
or, 0=(x-7)(x+16)
either,
0=x-7
or,7=x
x=7cm
Or,
0=x+16
or, -16=x
x= -16[impossible,as distance is never negative] so,
x=7cm
therefore,length = 7cm + 9 cm = 16cm and width = 7cm.
;)
let width(b)be x then,
length (l)=9cm+x
area =112 sq cm
now,
area of rectangle=l*b
or, 112=(9+x)x
or, 112=9x+x^2
or, 0=x^2+9x-112
or, 0=x^2+(16-7)x-112
or, 0=x^2+16x-7x-112
or, 0=x(x+16)-7(x+16)
or, 0=(x-7)(x+16)
either,
0=x-7
or,7=x
x=7cm
Or,
0=x+16
or, -16=x
x= -16[impossible,as distance is never negative] so,
x=7cm
therefore,length = 7cm + 9 cm = 16cm and width = 7cm.
;)
l = w + 9
lw = 112
Substitute l = w + 9 into the second equation
(w + 9)(w) = 112
w^2 + 9w - 112 = 0
(w+16)(w-7) = 0
w = -16
or
w = 7
we have to take the positive value, so w = 7
l = w + 9
l = 7 + 9
l = 16
The length is 16, and the width is 7
lw = 112
Substitute l = w + 9 into the second equation
(w + 9)(w) = 112
w^2 + 9w - 112 = 0
(w+16)(w-7) = 0
w = -16
or
w = 7
we have to take the positive value, so w = 7
l = w + 9
l = 7 + 9
l = 16
The length is 16, and the width is 7