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?

555AAABBBCCC

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the perpendicular lengths (AB,BC)
00:03 The perimeter of the triangle equals the sum of its sides
00:09 Substitute in the relevant values and calculate to express AB
00:22 Isolate AB
00:25 This is the expression for AB using BC
00:30 Apply the Pythagorean theorem
00:37 Substitute in the relevant values according to our given data and calculations
00:44 Open the parentheses
01:03 Arrange the equation
01:09 Divide by 2
01:21 Determine the 2 possibilities for BC
01:39 Substitute each option in the expression for AB
01:53 Each perpendicular can be either 3 or 4 corresponding to the other
01:58 This is 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?

555AAABBBCCC

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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