Question

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.

Answer 1

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

Step-by-step explanation:

Let W be the Width of rectangle

Then

According to "the length is 10 inches longer than its width"

L = W+10

Then according to formula of perimeter

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

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

Keywords: Perimeter, Linear Equations

Learn more about rectangles at:

• brainly.com/question/5424148
• brainly.com/question/5461619

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Answer 2

Answer:

$18in$ x $8 in$

Step-by-step explanation:

The area of a rectangle is:

$Area=width*length$

i will call the width $x$

$width=x$

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

$length=x+10$

thus, the area is

$Area =x(x+10)$

$Area=x^2+10x$

an the area according to the problem is equal to $144in^2$, so:

$144=x^2+10x\\0=x^2+10x-144$

Factoring the expression:

$(x+18)(x-8)=0$

clearing for $x$:

$x+18=0 --> x=-18$

and

$x-8=0-->x=8$

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

$18in$ x $8 in$

which when multiplied give the $144 in^2$ of area.