To prove that trapezoid ABCD is an isosceles trapezoid, we need to show that the non-parallel sides AD and BC are equal in length. According to your problem, the lengths of these sides are given as follows:
- AD = \(5y - 2\)
- BC = \(40 - y^2\)
To show that ABCD is isosceles, we set these two expressions equal to each other:
\[ 5y - 2 = 40 - y^2 \]
Now, we can rearrange this equation:
- Move all terms to one side:
\[ y^2 + 5y - 2 - 40 = 0 \]
- Combine like terms:
\[ y^2 + 5y - 42 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 5, c = -42 \).
First, we need to calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-42) = 25 + 168 = 193 \]
Now, we can substitute back into the quadratic formula:
\[ y = \frac{-5 \pm \sqrt{193}}{2 \cdot 1} = \frac{-5 \pm \sqrt{193}}{2} \]
Thus, the possible values of \( y \) are:
\[ y = \frac{-5 + \sqrt{193}}{2} \quad \text{or} \quad y = \frac{-5 - \sqrt{193}}{2} \]
Since \( \sqrt{193} \) is approximately 13.89, we can calculate:
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\( y \approx \frac{-5 + 13.89}{2} = \frac{8.89}{2} \approx 4.445 \) (this value is positive and feasible)
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\( y \approx \frac{-5 - 13.89}{2} = \frac{-18.89}{2} \approx -9.445 \) (this value is negative and would not be suitable for lengths)
Thus, the required value of \( y \) for ABCD to be an isosceles trapezoid is:
\[ \boxed{\frac{-5 + \sqrt{193}}{2}} \]
Which can be approximated as \( y \approx 4.445 \).