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

Angle A is equal to 30°.
Angle B is equal to 60°.
Angle C is equal to 90°.

Can these angles form a triangle?

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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