the area of one triangle is 9dm^2 and its perimeter is 4x-4. The area of a similar triangle is 4dm^2 and its perimeter is 2x+6. Find the perimeter of each triangle.

For similar triangles, the ratio of the perimeters must be the square root of the ratio of areas, which is (in this case) sqrt(9/4) = 3/2

Therefore 4x-4 = (3/2)(2x + 6) = 3x + 9
x = 13

The perimeter of the larger triangle is therefore 4x-4 = 48 dm. That of the smaller is 2x + 6 = 32 dm

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To find the perimeter of each triangle, we need to solve the given equations.

Let's consider the first triangle:
Area of the first triangle = 9 dm^2
Perimeter of the first triangle = 4x - 4

Now, let's consider the second triangle:
Area of the second triangle = 4 dm^2
Perimeter of the second triangle = 2x + 6

To find the perimeter of each triangle, we need to find the value of x.

Step 1: Set up the ratio of the areas of the two triangles:

(area of first triangle) / (area of second triangle) = (perimeter of first triangle)^2 / (perimeter of second triangle)^2

9 / 4 = (4x - 4)^2 / (2x + 6)^2

Step 2: Solve the equation:

(4x - 4)^2 / (2x + 6)^2 = 9 / 4

Cross-multiply:

4 * (4x - 4)^2 = 9 * (2x + 6)^2

16 * (4x - 4)^2 = 9 * (2x + 6)^2

Step 3: Solve the equation by taking the square root of both sides:

4x - 4 = ±√(9/16) * (2x + 6)

4x - 4 = ±(3/4) * (2x + 6)

Step 4: Expand and solve for x:

4x - 4 = ±(3/2)x + 9/2

Multiply through by 2 to eliminate the fraction:

8x - 8 = ±3x + 9

Step 5: Simplify and solve for x:

8x - 8 = 3x + 9 or 8x - 8 = -3x - 9

5x = 17 or 11x = -1

x = 17/5 or x = -1/11

Since the length cannot be negative, we discard x = -1/11.

Therefore, x = 17/5.

Step 6: Substitute the value of x into the given equations to find the perimeters of the triangles:

Perimeter of the first triangle = 4x - 4 = 4(17/5) - 4 = 68/5 - 20/5 = 48/5

Perimeter of the second triangle = 2x + 6 = 2(17/5) + 6 = 34/5 + 30/5 = 64/5

Therefore, the perimeter of the first triangle is 48/5 dm, and the perimeter of the second triangle is 64/5 dm.

To find the perimeters of each triangle, we need to determine the value of 'x' in the given equations.

Let's start with the first equation relating to the area and perimeter of the first triangle:

Area of the first triangle = 9 dm^2
Perimeter of the first triangle = 4x - 4

Now, the second equation relating to the area and perimeter of the second triangle:

Area of the second triangle = 4 dm^2
Perimeter of the second triangle = 2x + 6

To find the value of 'x', we can set up a proportion using the ratios of the areas and perimeters of the two triangles. Since the triangles are similar, their corresponding sides are in proportion.

Proportion of the areas:
Area of the first triangle / Area of the second triangle = (9 dm^2) / (4 dm^2)

Proportion of the perimeters:
Perimeter of the first triangle / Perimeter of the second triangle = (4x - 4) / (2x + 6)

Setting up the proportion of the areas, we have:

(9 dm^2) / (4 dm^2) = (4x - 4) / (2x + 6)

Now, we can solve for 'x' by cross-multiplying:

(9 dm^2) * (2x + 6) = (4 dm^2) * (4x - 4)

Expanding both sides of the equation:

18x + 54 = 16x - 16

Rearranging the equation:

18x - 16x = -16 - 54
2x = -70

Dividing both sides by 2 to solve for 'x':

2x / 2 = -70 / 2
x = -35

Now that we have found the value of 'x' as -35, we can substitute this value into the expressions for the perimeters of each triangle:

Perimeter of the first triangle = 4x - 4 = (4 * -35) - 4 = -136
Perimeter of the second triangle = 2x + 6 = (2 * -35) + 6 = -64

Therefore, the perimeter of the first triangle is -136 and the perimeter of the second triangle is -64.