The bottom of a rectangular box is 3 inches longer than it is wide. the diag is 15 inch. find areaofbottom box

Answer 1

Let, width of the bottom of the box = x
length of the bottom of box = x +3
Using pythgorian theorem,
x^2+(x+3)^2 = (15)^2
x^2+x^2+6x+9 = 225
2x^2+6x-216 = 0
(x+12)(x-9) = 0
x = -12 or x = 9
Since width cann't be negative, x = 9 is only valid answer.
Thus, width of the bottom of the box = 9 inch
Length of th bottom of the box = 9+3 = 12inch
Area = width*length = 9 *12 =108inch^2

Answer 2

This is really easy if you remember a 3-4-5 triangle. Multiply all these dimensions by 3 and you have a 9-12-15 triangle.

And 12 is three more than 9. So you simply have a 9 x 12 box with an area of 9 x 12 = 108 sq. in.

If you didn't know that, you would have to do it the long way using algebra and the pythagorean theorem.
Let w be the width.
Let w+3 be the length.

By the pythagorean theorem:
w² + (w+3)² = 15²

Simplifying:
w² + (w² + 6w + 9) = 225
2w² + 6w = 216

Divide both sides by 2:
w² + 3w = 108
w² + 3w - 108 = 0

Factoring:
(w + 12)(w - 9) = 0

So w = -12 or 9, but since it has to be positive,
w = 9

The length is 3 more than this:
l = 12

So if the width is 9 and the length is 12, the area is 9 x 12 or:
108 sq. in.

Answer 3

You should do your own homework, but I'll help you set up the equations to solve this.

Let's first label the lengths of the box's sides as x and y. Now, you are given the fact that one side is 3 inches longer than the other. That gives you an equation:

x = y+3

You are also told that the diagonal of the box is 15 inches. Since this is a rectangular box, the diagonal, along with two sides, form a right triangle, and so you can use the Pythagorean theorem (x^2 means "x squared"):

x^2 + y^2 = 15^2.

Now you can substitute the first equation into the second and solve for x and y:

(y+3)^2 + y^2 = 15^2
2y^2 + 6y + 9 = 225

I'll leave it to you to solve that, and then find x. Remember that the final quantity you need is the area of the rectangle, which is x*y.

*A word to others responding to the question*: You guys should not give away the answer. Homework *help* is okay, but solving it for someone completely is not -- they won't learn anything. And it's really disrespectful to their teachers, as well.

Answer 4

Using Pythagoras's theorem
15^2 = (w+3)^2 + w^2
225 = 2w^2 + 6w +9
2w^2 + 6w - 216 = 0
dividing by 2 yields;
w^2 + 3w - 108 = 0
(w-9)(w+12) =0
w = 9 OR w = -12
we cannot have a negative distance so the width (W) = 9
The other side is w + 3 = 12
AREA = 9 * 12 = 108 sq in

Answer 5

let x = width of box
x + 3 = length of box

x^2 + (x + 3)^2 = 15^2
x^2 + x^2 + 6x + 9 = 225
2x^2 + 6x - 216 = 0
x^2 + 3x - 108 = 0
(x + 12)(x - 9) = 0

x = 9, -12

The box is 9 inches wide and 12 inches long. Therefore, the area of the box is 9*12 = 108 square inches.

Answer 6

Let the length be x ,x+3
x^2+(x+3)^2 = 15^2 = 225
or
x^2+x^2+6x+9 = 225
or 2x^2+6x = 216
or x^2+3x = 108
orx^2+3x-108 =0
or(x+12)(x-9) =0

x = -12 or 9
-12 cannof be length
so x =9 and bottom = 12

Answer 7

It's 108 & the information regarding one side being 3 inches longer than the other ia not necessary because the only possible size for a rectangle with the diagonal=15 is a 9x12.

Answer 8

108 square inches
15 is the diagonal so it is the longest dimension of the bottom
we all know that 3,4 and 5 is a phythagorean triple so we will use this ratio to find the length and width. 3:4:5=9:12:15

Answer 9

Area = 36 units square

A = W x L
L = W + 3
W^2+(W+3)^2=15^2

Answer 10

rounded, the answer is approximately 139.12 inches squared

thats just a wild guess