Question

The area of a rectangular plank is 4500 cm². The plank was broken into two pieces, one of which is a square and the other a rectangle. Find the dimensions of the square if the length of the broken off rectangle is 120 cm.

Answer 1

Answer:

30cm

Step-by-step explanation:

Let the dimensions of the square is x cm.

The plank of area 4500 cm² was broken into two pieces, one of which is a square and the other a rectangle.

Length of the broken off rectangle = 120 cm

Width of the broken off rectangle = x cm

Length of plank = (120+x) cm

Width of plank = x cm

Area of a rectangle is

$Area=lenght\times width$

Area of plank is

$Area=(x+120)\times x$

$Area=x^2+120x$

The area of a rectangular plank is 4500 cm².

$x^2+120x=4500$

$x^2+120x-4500=0$

Splitting the middle term we get

$x^2+150x-30x-4500=0$

$x(x+150)-30(x+150)=0$

$(x+150)(x-30)=0$

Using zero product property we get

$(x+150)=0\Rightarrow x=-150$

$(x-30)=0\Rightarrow x=30$

x can not be a negative number because it is the side length of square. So, x=30.

Therefore, the side length of the square is 30cm.

Answer 2

Answer:

$Xs=30cm$ and $As=900 cm^{2}$

Step-by-step explanation:

At=As+Ar; A=b.h and $At=4500 cm^{2}$, then:  At=(x+120)x so

(x+120)x=4500, $x^{2} +120x-4500=0$ Applying cuadratic equation formula:$\frac{120+-\sqrt{120^{2} -4.1.-4500} }{2} =-\frac{120+-\sqrt{14400+18000} }{2}= x1=30 and x2=-150, finally Xs=30cm, and As=900cm^{2}$