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Part 1
The length of the base of an isosceles triangle is x. The length of a leg is 22xminus−77. The perimeter of the triangle is 2121. Find x. geometry

1 answer

To find the value of \( x \) for the given isosceles triangle, let's denote the lengths as follows:

  • The length of the base \( b = x \).
  • The length of each leg \( l = 22x - 77 \).

Since the triangle is isosceles, the perimeter \( P \) can be expressed as:

\[ P = b + 2l \]

Substituting the expressions for \( b \) and \( l \) into the perimeter equation:

\[ 2121 = x + 2(22x - 77) \]

Now, simplify the equation:

  1. Distribute \( 2 \) to \( (22x - 77) \): \[ 2121 = x + 44x - 154 \]

  2. Combine like terms: \[ 2121 = 45x - 154 \]

  3. Add \( 154 \) to both sides: \[ 2121 + 154 = 45x \] \[ 2275 = 45x \]

  4. Divide by \( 45 \): \[ x = \frac{2275}{45} \]

Now, simplify \( \frac{2275}{45} \). Calculate it efficiently:

  1. Find a common factor (if any):
    • Divide the numerator and denominator by \( 5 \): \[ x = \frac{455}{9} \]

Thus, the value of \( x \) is:

\[ x = \frac{455}{9} \]

This gives us the result, and we can also evaluate \( x \) as a decimal for practical purposes:

\[ x \approx 50.56 \]

Thus, the final answer for \( x \) is:

\[ \boxed{\frac{455}{9}} \] or approximately \( \boxed{50.56} \).