Triangle Side Lengths: Finding Legs When Perimeter is 12 cm

Right Triangle Legs with Perimeter Constraints

The perimeter of a triangle is 12 cm.

What are the lengths of its legs?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:06 Let's start by finding the perpendicular lengths: AB and BC .
00:11 Remember, the perimeter of a triangle is simply the sum of its si des.
00:16 Substitute the known values into our formula. Let's calculate to exp ress A B.
00:28 Now, isolate A B on one side of the equation.
00:32 Here is how A B is related to B C.
00:36 Next, use the Pythagorean theorem to help us along.
00:43 Substitute the given values based on our data and calculation s.
00:50 Now, let's open the parentheses step by step.
01:09 Rearrange the equation to make it easier to read.
01:15 We're going to divide everything by two.
01:27 Let's find the two possible values for B C.
01:45 Substitute each value into our expression for A B.
01:59 Each length can be either three or four, depending on the other.
02:05 And that's how we find the solution!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

The perimeter of a triangle is 12 cm.

What are the lengths of its legs?

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2

Step-by-step solution

This problem involves determining the lengths of the legs of a triangle whose perimeter is 12 cm, given that one side is 5 cm. To solve, consider the apparent context that implies a right triangle.

First, let's denote the three sides of the triangle as a a , b b , and c c , where c=5 c = 5 cm.

Considering the perimeter formula:

a+b+c=12 a + b + c = 12

Since c c is 5 cm, the equation becomes:

a+b+5=12 a + b + 5 = 12

Solving for a+b a + b :

a+b=7 a + b = 7

Assuming it is a right triangle with the side length of 5 cm as the hypotenuse:

c2=a2+b2 c^2 = a^2 + b^2

Where c=5 c = 5 , the equation is:

52=a2+b2 5^2 = a^2 + b^2
25=a2+b2 25 = a^2 + b^2

We need the integers a a and b b that satisfy both a+b=7 a + b = 7 and a2+b2=25 a^2 + b^2 = 25 .

To trial integer pairs from a+b=7 a + b = 7 :

- If a=3 a = 3 , then b=4 b = 4 .

Check a=3 a = 3 and b=4 b = 4 in the Pythagorean condition:

32+42=9+16=25 3^2 + 4^2 = 9 + 16 = 25

Hence, the pair satisfies both conditions.

Therefore, the lengths of the legs are 3cm 3 \, \text{cm} and 4cm 4 \, \text{cm} .

3

Final Answer

3 cm, 4 cm

Key Points to Remember

Essential concepts to master this topic
  • Perimeter Rule: Sum of all three sides equals total perimeter
  • Pythagorean Check: For legs 3 and 4: 32+42=9+16=25=52 3^2 + 4^2 = 9 + 16 = 25 = 5^2
  • Verification: Test both conditions: 3 + 4 + 5 = 12 and 32+42=52 3^2 + 4^2 = 5^2

Common Mistakes

Avoid these frequent errors
  • Ignoring the right triangle requirement
    Don't just find any two numbers that add to 7 like 1 + 6 or 2 + 5 = wrong triangle! These don't satisfy a2+b2=52 a^2 + b^2 = 5^2 , so they can't form a right triangle with hypotenuse 5. Always check both the perimeter condition AND the Pythagorean theorem.

Practice Quiz

Test your knowledge with interactive questions

Look at the triangle in the diagram. How long is side AB?

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FAQ

Everything you need to know about this question

How do I know which side is the hypotenuse?

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The hypotenuse is always the longest side in a right triangle. Since we're given one side is 5 cm and need to find two legs that add up to 7 cm (so each is less than 5), the 5 cm side must be the hypotenuse.

Why can't the legs be 1 cm and 6 cm?

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Let's check: 12+62=1+36=37 1^2 + 6^2 = 1 + 36 = 37 , but 52=25 5^2 = 25 . Since 37 ≠ 25, these lengths don't form a right triangle with hypotenuse 5 cm.

What if I get the legs backwards (4 cm and 3 cm)?

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That's perfectly fine! In a right triangle, it doesn't matter which leg you call 'first' or 'second'. Both 3 cm, 4 cm and 4 cm, 3 cm represent the same triangle.

How do I solve the system of equations?

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You have two equations:

  • a+b=7 a + b = 7 (from perimeter)
  • a2+b2=25 a^2 + b^2 = 25 (from Pythagorean theorem)

Try integer pairs from the first equation and check which one satisfies the second!

Are there other right triangles with perimeter 12?

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Yes! But this problem specifically shows a triangle with one side labeled as 5 cm. The 3-4-5 triangle is the only right triangle with integer sides and perimeter 12 that includes a side of length 5.

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