- Apply the triangle inequality \(x + (x+4) > 20\) to find \(x > 8\).
- Use the longest side constraint \(x+4 \le 20\) to find \(x \le 16\).
- Apply the acute triangle inequality \(x^2 + (x+4)^2 > 20^2\) to find \(x > 12\).
- Determine that the valid range for an acute triangle is \(12 < x \le 16\) and identify that \(14\) is the only option in this range.
- Final answer: \(\boxed{d}\).
Explanation
- Apply the Triangle Inequality Theorem and establish the range for x.
To determine the value of \(x\) that allows the lengths \(x\), \(x+4\), and \(20\) to form an acute triangle with \(20\) being the longest side, we start by applying fundamental geometric principles.
First, we apply the triangle inequality theorem. For any triangle with sides \(a, b, c\), the sum of the lengths of any two sides must be strictly greater than the length of the third side. Here, the side lengths are \(x\), \(x+4\), and \(20\). Given that \(20\) is the longest side, we must have:
\[x + (x + 4) > 20\]
\[2x + 4 > 20\]
\[2x > 16\]
\[x > 8\]
Next, we use the condition that \(20\) is the longest side, meaning the sum of the other two sides is at least as large as the longest side, or simply that no side can be longer than the longest side. Thus:
\[x + 4 \le 20\]
\[x \le 16\]
Combining these, we have the range \(8 < x \le 16\).
- Set up the acute triangle inequality condition.
For a triangle to be acute, the square of the length of the longest side must be strictly less than the sum of the squares of the lengths of the two shorter sides. Using the side lengths \(x\), \(x+4\), and \(20\), we set up the following inequality:
\[x^2 + (x+4)^2 > 20^2\]
Expanding the squared term \((x+4)^2 = x^2 + 8x + 16\), the expression becomes:
\[x^2 + x^2 + 8x + 16 > 400\]
\[2x^2 + 8x + 16 > 400\]
\[2x^2 + 8x - 384 > 0\]
Dividing the entire inequality by 2 simplifies it to:
\[x^2 + 4x - 192 > 0\]
- Solve the quadratic inequality.
To solve the quadratic inequality \(x^2 + 4x - 192 > 0\), we first find the roots of the corresponding quadratic equation \(x^2 + 4x - 192 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-192)}}{2(1)}\]
\[x = \frac{-4 \pm \sqrt{16 + 768}}{2}\]
\[x = \frac{-4 \pm \sqrt{784}}{2}\]
Since \(\sqrt{784} = 28\):
\[x = \frac{-4 \pm 28}{2}\]
The two roots are:
\[x_1 = \frac{-4 + 28}{2} = \frac{24}{2} = 12\]
\[x_2 = \frac{-4 - 28}{2} = \frac{-32}{2} = -16\]
For the quadratic \(x^2 + 4x - 192 > 0\), the solution lies outside the roots, so \(x > 12\) or \(x < -16\). Since \(x\) must be a positive length, we only consider \(x > 12\).
- Evaluate options and determine the final result.
Combining our previous conditions:
- From the triangle inequality and longest side constraint: \(8 < x \le 16\).
- From the acute triangle condition: \(x > 12\).
Intersecting these two intervals, we find that for the triangle to be acute, \(x\) must satisfy \(12 < x \le 16\). Looking at the options provided:
A\(10\) (Not in range)
B\(8\) (Not in range)
C\(12\) (Not in range, would result in a right triangle since \(12^2 + 16^2 = 144 + 256 = 400 = 20^2\))
D\(14\) (In range \(12 < 14 \le 16\))
Therefore, the only value from the options that results in an acute triangle is \(14\). The correct option is d.
Examples
In structural engineering, when designing roof trusses, it is essential to ensure that the triangular sections have specific angular properties for stability. If you are constructing a truss using wooden beams of length \(x\), \(x+4\), and a base support beam of \(20\) meters, you can use these inequalities to ensure the triangle remains acute. For example, by choosing \(x = 14\) meters, you guarantee that the truss creates an acute-angled triangle, which is often preferred for distributing weight effectively across the joints of the structure. This simple calculation prevents the design from becoming an obtuse structure which might be less efficient for specific roof pitches.