one side of an equilateral triangle is 7 dm shorter than one side of square.the sum of the perimeter of the two figure is 49 dm . find the perimeter of each figure.

To solve this problem, let's start by assigning variables to the unknowns.

Let's say the length of one side of the equilateral triangle is "x" dm. The length of one side of the square would then be "x + 7" dm.

The perimeter of the equilateral triangle would be 3 times the length of one side, so it would be 3x dm.
The perimeter of the square would be 4 times the length of one side, so it would be 4(x + 7) dm.

According to the problem, the sum of the perimeters of both figures is 49 dm.

So we can set up the following equation to represent the problem:

3x + 4(x + 7) = 49

Now, let's solve for x:

3x + 4x + 28 = 49
7x + 28 = 49
7x = 49 - 28
7x = 21
x = 21 / 7
x = 3

We have found that the length of one side of the equilateral triangle is 3 dm.

Now, let's calculate the perimeter of each figure:

Perimeter of the equilateral triangle = 3x dm = 3 * 3 dm = 9 dm
Perimeter of the square = 4(x + 7) dm = 4 * (3 + 7) dm = 4 * 10 dm = 40 dm

Therefore, the perimeter of the equilateral triangle is 9 dm and the perimeter of the square is 40 dm.

x = length of one side of a square

x - 7 = length of one side of an equilateral triangle

Equation:
4x + 3(x - 7) = 49

Solving for x:
4x + 3x - 21 = 49
7x - 21 = 49
7x = 70
x = 10

Now you should be able to determine the perimeter of each figure.

I hope this helps.