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.

(x+120)x=4500, [tex]x^{2} +120x-4500=0[/tex] Applying cuadratic equation formula:[tex]\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}[/tex]

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erinna erinna · Answer 2 · 2019-12-03T08:08:55+01:00

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

[tex]Area=lenght\times width[/tex]

Area of plank is

[tex]Area=(x+120)\times x[/tex]

[tex]Area=x^2+120x[/tex]

The area of a rectangular plank is 4500 cm².

[tex]x^2+120x=4500[/tex]

[tex]x^2+120x-4500=0[/tex]

Splitting the middle term we get

[tex]x^2+150x-30x-4500=0[/tex]

[tex]x(x+150)-30(x+150)=0[/tex]

[tex](x+150)(x-30)=0[/tex]

Using zero product property we get

[tex](x+150)=0\Rightarrow x=-150[/tex]

[tex](x-30)=0\Rightarrow x=30[/tex]

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.