A rectangle has an area of 260 square inches and a width which is six inches more than twice the length. Find the length and the width of the rectangle.

Answer 1

I solved this in the whiteboard session. Please let me know if it's accessible to you there. It won't let me upload a screenshot so I'll describe what I did there.

Step 1: Draw and label your diagram.

Draw a rectangle. Label your width and label your height. I labeled the width as the longer side.

Label the "length" side as "x". We do this because we don't know what the length is.

Label your width side: "6in plus twice the length"...meaning two times "x"...so 6in + 2x

Step 2: Setup your equation

We want to solve for x. To do this, we must remember that area=length times width

Therefore, we multiple "x" and "6in +2x" as so:

(6x+2x^2) Protip: You can temporarily leave out the "in". It won't affect your calculation because it's just the "name" to your measurement, inches.

Step 3: Set it equal to 260 and solve for x

6x+2x^2 = 260 \\ I'm going to skip this algebra step to save time but if you don't know how to do it, please comment so I can elaborate further.

You should get x=10 or -13

Because you can't have a "negative length", x = 10.

So now, plug that x value into your "side equations" to get a width of: 26in and a length of 10in

Answer 2

For this problem, we will use the area formula for a rectangle which is A = l x w, only we will need to represent both dimensions using only one variable. Since the width is described in terms of the length, let's let x represent the length, and then we will let 2x + 6 represent the width. Note: this is an odd situation where the width is going to be the longer of the two measurements. Using the given area of 260, the equation will look like this: 260 = x(2x + 6). Eliminating the ( ), we have 260 = 2x² + 6x.

If we move all the terms to one side, and then divide everything by 2, the modified equation becomes x² + 3x - 130 = 0. Thankfully, this is a factorable trinomial which factors into ( x - 10)(x + 13 ) = 0. By the Zero Product Property, x = 10 or x = -13. Since a negative measurement for the length doesn't make sense, we can only accept x = 10. So the length of the rectangle is 10, and by the expression 2x + 6, we have the width = 26. By multiplying these two dimensions, we get an area of 260 which confirms that we have solved correctly.

Answer 3

Hey! Lets solve this shall we ?

So, we must find the length and width of this rectangle given the Area of A = 260 square inches. We yield the formula of Area of a Rectangle such that A = length * width. If the width is 6 inches more than 2 times the length, we have w = 2x + 6, where the length is l = x. Now lets plug everything in to get

A = l*w

260 = (x)(2x+6)............We distribute the x inside to get

260 = 2x² + 6x............Then we subtract 260 to the right side to obtain

0 = 2x² + 6x - 260..........We factor out a 2 to obtain

0 = 2(x² + 3x - 130).........Divide the 2 to both sides to remain

x² + 3x - 130 = 0..........Now we factor this into

(x+13)(x-10) = 0.....Then we apply the Zero-Product property of Equality such that

x + 13 = 0; x - 10 = 0.......Then solve for x

x = -13; x = 10........We cant have a negative length so we use x = 10 inches for the length and

width = 2(10) + 6 = 26 inches

I hope this helped!