Let's denote the length of the longer leg of the right triangle as \( x \). Therefore, the length of the shorter leg will be \( x - 7 \).
According to the Pythagorean theorem, the relationship between the legs and the hypotenuse \( c \) in a right triangle is given by:
\[ c^2 = (x - 7)^2 + x^2 \]
We want to find when the hypotenuse \( c \) is at least 17 inches long:
\[ c \geq 17 \]
Squaring both sides (since both sides are positive), we get:
\[ c^2 \geq 17^2 \]
\[ c^2 \geq 289 \]
Substituting \( c^2 \) from the Pythagorean theorem gives us:
\[ (x - 7)^2 + x^2 \geq 289 \]
Expanding the left-hand side, we have:
\[ (x^2 - 14x + 49) + x^2 \geq 289 \]
Combining like terms:
\[ 2x^2 - 14x + 49 \geq 289 \]
Subtracting 289 from both sides:
\[ 2x^2 - 14x + 49 - 289 \geq 0 \]
Simplifying gives:
\[ 2x^2 - 14x - 240 \geq 0 \]
Dividing through by 2 for simplicity:
\[ x^2 - 7x - 120 \geq 0 \]
Next, let's find the roots of the quadratic equation \( x^2 - 7x - 120 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -7, c = -120 \):
\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1} \] \[ x = \frac{7 \pm \sqrt{49 + 480}}{2} \] \[ x = \frac{7 \pm \sqrt{529}}{2} \] \[ x = \frac{7 \pm 23}{2} \]
Calculating the roots:
\[ x = \frac{30}{2} = 15 \quad \text{and} \quad x = \frac{-16}{2} = -8 \]
The roots are \( x = 15 \) and \( x = -8 \).
Now, to find the intervals where \( x^2 - 7x - 120 \geq 0 \), we test intervals based on the roots:
- \( (-\infty, -8) \)
- \( (-8, 15) \)
- \( (15, \infty) \)
By testing a point from each interval:
- For \( x < -8 \) (e.g., \( x = -9 \)): \( (-9)^2 - 7(-9) - 120 = 81 + 63 - 120 = 24 \geq 0 \) (true)
- For \( -8 < x < 15 \) (e.g., \( x = 0 \)): \( 0^2 - 7(0) - 120 = -120 < 0 \) (false)
- For \( x > 15 \) (e.g., \( x = 16 \)): \( 16^2 - 7(16) - 120 = 256 - 112 - 120 = 24 \geq 0 \) (true)
Thus, the solution for the inequality \( x^2 - 7x - 120 \geq 0 \) is:
\[ x \leq -8 \quad \text{or} \quad x \geq 15 \]
Since we are talking about lengths, we discard the solution \( x \leq -8 \).
So the only valid solution is \( x \geq 15 \).
Therefore, the hypotenuse is at least 17 inches long when the longer leg is greater than or equal to 15 inches.
The correct answer is:
When the longer leg is greater than or equal to 15 inches.
1 answer