The length of a rectangle is 5cm more than it's width. If the width is increased by 2 cm and the length is increased by 3 cm a new rectangle is formed that has an area of 46 cm squared more than the original rectangle. Find dimensions of the original rectangle.

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calculista calculista · Answer 1 · 2020-02-22T12:18:15+01:00

Answer:

The dimensions of the original rectangle are:

Length 11 cm

Width 6 cm

Step-by-step explanation:

Let

x ----> the original length

y ----> the original width

we know that

The original area is

[tex]A_1=xy[/tex] ----> equation A

[tex]x=y+5[/tex] ----> equation B

The new area is

[tex]A_2=(x+3)(y+2)[/tex] ----> equation C

[tex]A_2=A_1+46[/tex] ----> equation D

substitute equation A and equation C in equation D

[tex](x+3)(y+2)=xy+46[/tex] ----> equation E

substitute equation B in equation E

[tex](y+5+3)(y+2)=(y+5)y+46[/tex]

solve for y

([tex]y+8)(y+2)=y^2+5y+46[/tex]

[tex]y^2+10y+16=y^2+5y+46[/tex]

[tex]10y-5y=46-16\\5y=30\\y=6\ cm[/tex]

Find the value of x

[tex]x=6+5=11\ cm[/tex]

therefore

The dimensions of the original rectangle are:

Length 11 cm

Width 6 cm