- Define the isosceles triangle with sides \(Q, Q\) and vertex angle \(45^{\circ}\).
- Use the Law of Cosines: \(b^2 = Q^2 + Q^2 - 2Q^2 \cos(45^{\circ})\).
- Substitute \(\cos(45^{\circ}) = \frac{\sqrt{2}}{2}\) and simplify: \(b^2 = Q^2(2 - \sqrt{2})\).
- Take the square root to obtain the base length: \(\boxed{Q \sqrt{2 - \sqrt{2}}}\).
Explanation
- Define the triangle parameters and objective.
An isosceles triangle has two equal sides of length \(Q\) and a vertex angle of \(45^{\circ}\). Let the triangle be \(ABC\), where \(AB = AC = Q\) and \(\angle BAC = 45^{\circ}\). We want to find the length of the base \(BC\). Let this length be \(b\).
- Apply the Law of Cosines to solve for the base.
The Law of Cosines relates the side lengths of a triangle to the cosine of one of its angles. For triangle \(ABC\), the side \(BC\) opposite to \(\angle A\) is given by the formula:
\[b^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(A)\]
Substituting the known values \(AB = Q\), \(AC = Q\), and \(\angle A = 45^{\circ}\):
\[b^2 = Q^2 + Q^2 - 2(Q)(Q) \cos(45^{\circ})\]
Since \(\cos(45^{\circ}) = \frac{\sqrt{2}}{2}\), we have:
\[b^2 = 2Q^2 - 2Q^2 \left( \frac{\sqrt{2}}{2} \right)\]
\[b^2 = 2Q^2 - Q^2 \sqrt{2}\]
\[b^2 = Q^2(2 - \sqrt{2})\]
- Calculate the length of the base and identify the result.
Taking the square root of both sides to find \(b\):
\[b = \sqrt{Q^2(2 - \sqrt{2})} = Q \sqrt{2 - \sqrt{2}}\]
This matches one of the provided options.
Examples
This geometric calculation is fundamental in engineering and design, such as when determining the span of a roof truss or the length of a brace in a structure. For instance, if you are building an A-frame structure where two equal beams of length 10 feet meet at a \(45^{\circ}\) angle, you can use the Law of Cosines to find the exact width required for the foundation, ensuring the structure is built to the correct proportions. Similar calculations apply in robotics when finding the distance between two joints in a robotic arm configuration.