Area of a Rectangle: Using variables

Let's multiply side AB by side BC

We'll set up the data as follows:

$5(2x+4)=40$

Let's multiply 5 by each term in parentheses:

$10x+20=40$

We'll move 20 to the right side and change its sign accordingly:

$10x=40-20$

Now we get:

$10x=20$

Let's divide both sides by 10:

$x=2$

The area of the rectangle is equal to the length multiplied by the width.

Let's begin by presenting the known data:

$48=4\times2x$

$48=8x$

Lastly let's divide both sides by 8:

$x=6$

The area of a rectangle is equal to the length multiplied by the width.

Let's insert the known data into the formula:

$27=3\times3x$

$27=9x$

Lastly let's divide both sides by 9:

$x=3$

The area of the rectangle is equal to the length multiplied by the width.

Let's present the known data:

$24=3\times(2x+2)$

$24=6x+6$

We'll move 6 to the left side and maintain the appropriate sign:

$24-6=6x$

$18=6x$

We'll divide both sides by 6:

$x=3$

The area of the rectangle is equal to the length multiplied by the width.

Let's present the known data:

$28=4\times(x+5)$

$28=4x+20$

We'll move 20 to the left side and maintain the appropriate sign:

$28-20=4x$

$8=4x$

Let's divide both sides by 4:

$x=2$

The area of the rectangle is equal to its length multiplied by its width.

We begin by inserting the given data into this formula:

$30=3\times2x$

$30=6x$

Lastly we divide both sides by 6:

$x=5$

The area of the rectangle is equal to the length multiplied by the width.

Let's begin by presenting the known data:

$50=5\times4x$

$50=20x$

Let's finish by dividing both sides by 20:

$x=2.5$

The area of a rectangle equals length times width

Let's put the data into the formula:

$S=x\times\frac{x}{2}=\frac{x^2}{2}$

Since we are given that x equals 4, let's substitute it into the formula accordingly:

$S=\frac{4^2}{2}=\frac{16}{2}=8$

The area of a rectangle is equal to its length multiplied by its width.

Let's first input the known data into the formula:

$32=8x$

Let's now reduce both sides of the equation by the (HCF) highest common factor 8:

$x=4$

The area of a rectangle is equal to length multiplied by width

Let's input the known data into the formula:

$S=x\times x^2$

Since we are given that x equals 9, let's substitute it into the formula:

$S=9\times9^2=9\times81=729$

The area of a rectangle is equal to its length multiplied by its width.

Let's begin by inserting the known data into the formula:

$72=x\times2x$

$72=2x^2$

Let's proceed to simplify both sides of the equation by the HCF (highest common factor ) in this case 2:

$36=x^2$

Finally we remove the square root in order to solve the equation as follows:

$x=6$

The area of a rectangle is equal to its length multiplied by its width.

Let's begin by inserting the known data into the formula as follows :

$72=2x\times4x$

$72=8x^2$

Let's proceed to simplify both sides of the equation by the (HCF) the highest common factor, in this case 8 :

$9=x^2$

Finally let's remove the square root:

$x=3$

We know the area of DBHF and also the length of HF

We will substitute into the formula in order to find BF, let's call the side BF as X:

$4\times x=20$

We'll divide both sides by 4:

$x=5$

Therefore, BF equals 5

Now we can calculate the volume of the box:

$8\times4\times5=32\times5=160$

Based on the given data, we can argue that:

$BD=HF=2$

We know the area of ABCD and also the length of DB

We'll substitute in the formula to find CD, let's call the side CD as X:

$2\times x=12$

We'll divide both sides by 2:

$x=6$

Therefore, CD equals 6

Now we can calculate the volume of the box:

$6\times2\times3=12\times3=36$

Since we are given the area of rectangle CAEG and length AE, we can find GE:

Let's denote GE as X and substitute the data in the rectangle area formula:

$3\times x=15$

Let's divide both sides by 3:

$x=5$

Therefore GE equals 5

Since we are given the area of rectangle ABFE and length AE, we can find EF:

Let's denote EF as Y and substitute the data in the rectangle area formula:

$3\times y=24$

Let's divide both sides by 3:

$y=8$

Therefore EF equals 8

Now we can calculate the volume of the box:

$3\times5\times8=15\times8=120$

The area of a rectangle is equal to length multiplied by width

Let's input the known data into the formula:

$S=2x\times(2x-8)$

$S=2x\times2x-2x\times8$

$S=4x^2-16x$

The area of the rectangle is equal to the length times the width:

$S=x\times(x-4)$

$S=x^2-4x$

The area of the rectangle is equal to the length multiplied by the width.

Let's list the known data:

$22x=\frac{1}{2}x\times(x+8)$

$22x=\frac{1}{2}x^2+\frac{1}{2}x8$

$22x=\frac{1}{2}x^2+4x$

$0=\frac{1}{2}x^2+4x-22x$

$0=\frac{1}{2}x^2-18x$

$0=\frac{1}{2}x(x-36)$

For the equation to be equal, x needs to be equal to 36

The area of the rectangle equals:

$S=AB\times BC$

$64=AB\times X$

Since it is given that side AB is 4 times larger than side BC

We can state that:

$AB=4BC=4X$

Now let's substitute this information into the formula for calculating the area:

$64=4x\times x$

$64=4x^2$

Let's divide both sides by 4:

$16=x^2$

We'll take the square root and get:

$4=x$

In other words, BC equals 4

Let us begin by reminding ourselves of the formula to calculate the area of a rectangle: width X length

$S=w⋅h$

When:

S = area

w = width

h = height

We take data from the sides of the rectangle in the figure.$w=b+8$$h=a+3$

We then substitute the above data into the formula in order to calculate the area of the rectangle:

$S=w⋅h = (b+8)(a+3)$

We use the formula of the extended distributive property:

$(a+b)(c+d)=ac+ad+bc+bd$

We substitute once more and solve the problem as follows:

$S=(b+8)(a+3)=(b)(a)+(b)(3)+(8)(a)+(8)(3)$

$(b)(a)+(b)(3)+(8)(a)+(8)(3)=ab+3b+8a+24$

Therefore, the correct answer is option B: ab+8a+3b+24.

Keep in mind that, since there are only addition operations, the order of the terms in the expression can be changed and, therefore,

$ab+3b+8a+24=ab+8a+3b+24$