What this formula means in simpler terms is that to find the perimeter of a triangle, you just add together the lengths of each of its 3 sides.

What this formula means in simpler terms is that to find the perimeter of a triangle, you just add together the lengths of each of its 3 sides.

What this formula means in simpler terms is that to find the perimeter of a triangle, you just add together the lengths of each of its 3 sides.

What this formula means in simpler terms is that to find the perimeter of a triangle, you just add together the lengths of each of its 3 sides.

What this formula means in simpler terms is that to find the perimeter of a triangle, you just add together the lengths of each of its 3 sides.

What this formula means in simpler terms is that to find the perimeter of a triangle, you just add together the lengths of each of its 3 sides.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

This particular example is called an equilateral triangle, because all three sides are of equal length. But remember that the perimeter formula is the same for any kind of triangle.

In another example, where a = 4, b = 3, and c=5, the perimeter would be: P = 3 + 4 + 5, or 12.

In another example, where a = 4, b = 3, and c=5, the perimeter would be: P = 3 + 4 + 5, or 12.

In another example, where a = 4, b = 3, and c=5, the perimeter would be: P = 3 + 4 + 5, or 12.

In this example, the side lengths are each 5cm, so the correct value for the perimeter is 15cm.

If, for example, you know that side a = 3 and side b = 4, then plug those values into the formula as follows: 32 + 42 = c2. If you know the length of side a = 6, and the hypotenuse c = 10, then you should set the equation up like so: 62 + b2 = 102.

If, for example, you know that side a = 3 and side b = 4, then plug those values into the formula as follows: 32 + 42 = c2. If you know the length of side a = 6, and the hypotenuse c = 10, then you should set the equation up like so: 62 + b2 = 102.

In the first example, square the values in 32 + 42 = c2 and find that 25= c2. Then calculate the square root of 25 to find that c = 5. In the second example, square the values in 62 + b2 = 102 to find that 36 + b2 = 100. Subtract 36 from each side to find that b2 = 64, then take the square root of 64 to find that b = 8.

In our first example,P = 3 + 4 + 5, or 12. In our second example, P = 6 + 8 + 10, or 24.

In our first example,P = 3 + 4 + 5, or 12. In our second example, P = 6 + 8 + 10, or 24.

In our first example,P = 3 + 4 + 5, or 12. In our second example, P = 6 + 8 + 10, or 24.

In our first example,P = 3 + 4 + 5, or 12. In our second example, P = 6 + 8 + 10, or 24.

In our first example,P = 3 + 4 + 5, or 12. In our second example, P = 6 + 8 + 10, or 24.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

For example, imagine a triangle with side lengths 10 and 12, and an angle between them of 97°. We will assign variables as follows: a = 10, b = 12, C = 97°.

c2 = 102 + 122 - 2 × 10 × 12 × cos(97). c2 = 100 + 144 – (240 × -0. 12187) (Round the cosine to 5 decimal places. ) c2 = 244 – (-29. 25) c2 = 244 + 29. 25 (Carry the minus symbol through when cos(C) is negative!) c2 = 273. 25 c = 16. 53

c2 = 102 + 122 - 2 × 10 × 12 × cos(97). c2 = 100 + 144 – (240 × -0. 12187) (Round the cosine to 5 decimal places. ) c2 = 244 – (-29. 25) c2 = 244 + 29. 25 (Carry the minus symbol through when cos(C) is negative!) c2 = 273. 25 c = 16. 53

c2 = 102 + 122 - 2 × 10 × 12 × cos(97). c2 = 100 + 144 – (240 × -0. 12187) (Round the cosine to 5 decimal places. ) c2 = 244 – (-29. 25) c2 = 244 + 29. 25 (Carry the minus symbol through when cos(C) is negative!) c2 = 273. 25 c = 16. 53

c2 = 102 + 122 - 2 × 10 × 12 × cos(97). c2 = 100 + 144 – (240 × -0. 12187) (Round the cosine to 5 decimal places. ) c2 = 244 – (-29. 25) c2 = 244 + 29. 25 (Carry the minus symbol through when cos(C) is negative!) c2 = 273. 25 c = 16. 53

c2 = 102 + 122 - 2 × 10 × 12 × cos(97). c2 = 100 + 144 – (240 × -0. 12187) (Round the cosine to 5 decimal places. ) c2 = 244 – (-29. 25) c2 = 244 + 29. 25 (Carry the minus symbol through when cos(C) is negative!) c2 = 273. 25 c = 16. 53

c2 = 102 + 122 - 2 × 10 × 12 × cos(97). c2 = 100 + 144 – (240 × -0. 12187) (Round the cosine to 5 decimal places. ) c2 = 244 – (-29. 25) c2 = 244 + 29. 25 (Carry the minus symbol through when cos(C) is negative!) c2 = 273. 25 c = 16. 53

In our example: 10 + 12 + 16. 53 = 38. 53, the perimeter of our triangle!

In our example: 10 + 12 + 16. 53 = 38. 53, the perimeter of our triangle!

In our example: 10 + 12 + 16. 53 = 38. 53, the perimeter of our triangle!

In our example: 10 + 12 + 16. 53 = 38. 53, the perimeter of our triangle!

In our example: 10 + 12 + 16. 53 = 38. 53, the perimeter of our triangle!