The perimeters of the triangles shown are equal. Find the side lengths of each triangle.

A
B
C
4x+4
4x+3
4x
4x
4x
5x+5
P
Q
R
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Part 1
AB overbarequals
  
enter your response here
PQ overbarequals
  
enter your response here
BC overbarequals
  
enter your response here
QR overbarequals
  
enter your response here
AC overbarequals
  
enter your response here
PR overbarequals
  
enter your response here
​(Type a whole​ number.)

1 answer

To solve for the side lengths of each triangle, we start by setting the perimeters for both triangles \(ABC\) and \(PQR\) equal to each other, since they are given to be equal.

Step 1: Find the Perimeter of Triangle ABC

The sides of triangle \(ABC\) are:

  • \(AB = 4x + 4\)
  • \(BC = 4x\)
  • \(AC = 4x + 3\)

The perimeter of triangle \(ABC\) is given by: \[ \text{Perimeter of } ABC = AB + BC + AC = (4x + 4) + (4x) + (4x + 3) \] Combine the terms: \[ \text{Perimeter of } ABC = 12x + 7 \]

Step 2: Find the Perimeter of Triangle PQR

The sides of triangle \(PQR\) are:

  • \(PQ = 4x\)
  • \(QR = 5x + 5\)
  • \(PR = 4x\)

The perimeter of triangle \(PQR\) is given by: \[ \text{Perimeter of } PQR = PQ + QR + PR = (4x) + (5x + 5) + (4x) \] Combine the terms: \[ \text{Perimeter of } PQR = 13x + 5 \]

Step 3: Set the Perimeters Equal

Since the perimeters of the triangles are equal, we have: \[ 12x + 7 = 13x + 5 \]

Step 4: Solve for \(x\)

Subtract \(12x\) from both sides: \[ 7 = x + 5 \]

Subtract \(5\) from both sides: \[ x = 2 \]

Step 5: Find the Side Lengths

Now, substitute \(x = 2\) back into the expressions for the side lengths.

For triangle \(ABC\):

  • \(AB = 4(2) + 4 = 8 + 4 = 12\)
  • \(BC = 4(2) = 8\)
  • \(AC = 4(2) + 3 = 8 + 3 = 11\)

For triangle \(PQR\):

  • \(PQ = 4(2) = 8\)
  • \(QR = 5(2) + 5 = 10 + 5 = 15\)
  • \(PR = 4(2) = 8\)

Final Result

Thus, the side lengths are:

  • For triangle \(ABC\):

    • \(AB = 12\)
    • \(BC = 8\)
    • \(AC = 11\)
  • For triangle \(PQR\):

    • \(PQ = 8\)
    • \(QR = 15\)
    • \(PR = 8\)

Here are the requested results:

  • \(AB = 12\)
  • \(PQ = 8\)
  • \(BC = 8\)
  • \(QR = 15\)
  • \(AC = 11\)
  • \(PR = 8\)