A piece of wire x cm long is to be cut into two pieces, each to bent to be a square. the length of a side of one square is to be 9 times the length of a side of the other . Express the sum of the areas of two squares in term of x .

I think you read "nine times" as "nine tenths". Also, the sum of the perimeters is x. In that case, the two perimeters are 9/10 x and 1/10 x. So, the areas sum to

((9/10 x)/4)^2 + ((1/10 x)/4)^2 = 41/800 x^2

Let's denote the length of a side of one square as "a" and the length of a side of the other square as "b". We are given that the total length of the wire, x cm, needs to be used to create these two squares.

Now, we can create two equations based on the given information:

1. Perimeter equation: Since the wire is cut into two pieces, each piece will form the perimeter of a square. Therefore,

2a + 2b = x (equation 1)

2. Side length relation: It is given that one side is 9 times the length of the other side. So,

a = 9b (equation 2)

We have two equations (equation 1 and equation 2) with two variables (a and b). We can solve these equations simultaneously to find the values of a and b.

From equation 1, we can write:

2a + 2b = x

Substituting the value of a from equation 2:

2(9b) + 2b = x

18b + 2b = x

20b = x

b = x/20

Substituting the value of b back into equation 2, we get:

a = 9b

a = 9(x/20)

Simplifying, we get:

a = 9x/20

Now, we can find the area of each square using the formulas:

Area of a square = (side length)^2

Area of the square with side length "a" = (9x/20)^2 = 81x^2/400

Area of the square with side length "b" = (x/20)^2 = x^2/400

Finally, the sum of the areas of the two squares is:

81x^2/400 + x^2/400

Simplifying, we obtain the sum of the areas in terms of x:

(82x^2)/400

Hence, the sum of the areas of the two squares in terms of x is (82x^2)/400.

To solve this problem, let's break it down step by step:

1. We have a piece of wire that is x cm long.
2. We need to cut the wire into two pieces.
3. Each piece will be bent to form a square.
4. The length of one square's side will be 9 times the length of the other square's side.
5. We need to express the sum of the areas of the two squares in terms of x.

Let's denote the length of the smaller square's side as s. Then, the length of the larger square's side will be 9s, as given in the problem.

To find the value of s, we can consider the perimeter of the squares. The perimeter is equal to the length of the wire, which is x cm.

The perimeter of a square is given by P = 4s. So for the smaller square, its perimeter is 4s, and for the larger square, it is 4(9s) = 36s.

Since we are cutting the wire into two pieces, the sum of the perimeters of the squares will be equal to the length of the wire:

Perimeter of smaller square + Perimeter of larger square = x cm
4s + 36s = x cm

Combining like terms, we get:
40s = x

To find the value of s in terms of x, we divide both sides of the equation by 40:
s = x/40

Now that we have the value of s, we can find the areas of the squares. Recall that the area of a square is given by A = s^2.

The area of the smaller square is (x/40)^2, and the area of the larger square is (9s)^2 = (9(x/40))^2.

The sum of the areas of the two squares is:
A = (x/40)^2 + (9(x/40))^2

Simplifying this expression further, we get:
A = (x^2/1600) + (81x^2/1600)
A = (x^2 + 81x^2)/1600
A = (82x^2)/1600

Therefore, the sum of the areas of the two squares in terms of x is (82x^2)/1600.

perimeter of new squares sum=x

so one length is 9/10 x, the other x

Then

area sum= (.9x/4)^2+(.1x/4)^2

= 1/16( .81x^2+.01x^2)

=1/16( .82x^2)