Recall the formula for finding the area of a rectangle. Define a variable for the width and set up an equation to find the dimensions of a rectangle that has an area of 144 square inches, given that the length is 10 inches longer than its width.

[tex]2L+2W=144\\2(W+10)+2W=144\\2W+20+2W=144\\4W+20=144\\Subtracting\ 20\ from\ both\ sides\\4W+20-20=144-20\\4W=124\\Dividing\ both\ sides\ by\ 4\\\frac{4W}{4}=\frac{124}{4}\\W=31\ inches\\L=W+10 = 31+10 = 41\ inches[/tex]

The length of rectangle is 41 inches and width is 31 inches

Keywords: Perimeter, Linear Equations

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rocioo rocioo · Answer 2 · 2020-02-20T04:56:06+01:00

Answer:

[tex]18in[/tex] x [tex]8 in[/tex]

Step-by-step explanation:

The area of a rectangle is:

[tex]Area=width*length[/tex]

i will call the width [tex]x[/tex]

[tex]width=x[/tex]

and since the length is 10 inches longer than the width :

[tex]length=x+10[/tex]

thus, the area is

[tex]Area =x(x+10)[/tex]

[tex]Area=x^2+10x[/tex]

an the area according to the problem is equal to [tex]144in^2[/tex], so:

[tex]144=x^2+10x\\0=x^2+10x-144[/tex]

Factoring the expression:

[tex](x+18)(x-8)=0\\[/tex]

clearing for [tex]x[/tex]:

[tex]x+18=0 --> x=-18[/tex]

and

[tex]x-8=0-->x=8[/tex]

since we are talking about sides of a rectangle we can ignore the negative sign in -18, and the dimensions of the rectangle are:

[tex]18in[/tex] x [tex]8 in[/tex]

which when multiplied give the [tex]144 in^2[/tex] of area.