Question

Solve geometrically the equation \(\frac{9}{x}=\frac{x}{5}\) by beginning with a semicircle of diameter \(9+5=14\)

Short answer

Answer: The solution for the geometric equation \(\frac{9}{x}=\frac{x}{5}\) is \(x = 3\sqrt{5}\).

Step-by-step solution

Draw the semicircle of diameter 14

Draw a semicircle with diameter \(14\), which will be the line segment \(AB\). Let the center of the circle be \(O\).

Define the given lengths on the diameter

Mark points \(C\) and \(D\) on the diameter such that \(AC = 9\) and \(CD = 5\).

Draw the perpendiculars

Draw the perpendiculars from points \(C\) and \(D\) to the semicircle, meeting the semicircle at points \(E\) and \(F\), respectively.

Analyze the triangles

Observe the triangles \(AEC\) and \(DFA\). We have \(\angle AEC = \angle DFA = 90°\) because they are radii perpendicular to the tangent on a circle. Moreover, \(\angle CAE = \angle FDA\) because they are both inscribed angles in the same semicircle. Therefore, \(\triangle AEC \sim \triangle DFA\) by the AA similarity criterion.

Set up the proportion and solve

Since \(\triangle AEC \sim \triangle DFA\), we can write the following proportions with respect to corresponding sides: \(\frac{AE}{CE} = \frac{DF}{FA}\). From the given equation, we have \(\frac{9}{x} = \frac{x}{5}\), and now we can set the lengths as follows: \(CE = x\), \(AE=9\), \(DF=x\), and \(FA=5\). So we have \(\frac{9}{x}=\frac{x}{5}\). Now, solving this equation for \(x\), we get \(x^2=45\). Taking the square root of both sides, we find \(x = 3\sqrt{5}\).

Conclusion

The solution for the geometric equation \(\frac{9}{x}=\frac{x}{5}\) is \(x = 3\sqrt{5}\) which can be obtained by finding similar triangles within a semicircle of diameter \(14\). This geometric interpretation gives us both a visual representation and a unique way of solving the problem.

Key concepts

Semicircle Geometry

A semicircle is half of a circle created by cutting a whole circle along its diameter. In the given problem, the semicircle is drawn with a diameter of 14. This is because the sum of the numbers in the equation to be solved (\(\frac{9}{x} = \frac{x}{5}\)) is 14. The semicircle here helps set the stage for solving the given equation using geometric concepts.
Drawing a semicircle with a specific diameter provides a tangible geometric representation, where the diameter forms the base of the semicircle. This line segment aids in visualizing and constructing other geometric elements, such as triangles. For our exercise, a semicircle of diameter 14 allows us to place significant points on the semicircle that will help solve the equation geometrically.
By dividing the diameter into segments corresponding to the given lengths, we can start building further geometric shapes that are crucial for the problem-solving process. Here, points \(C\) and \(D\) divide the diameter into lengths of 9 and 5 respectively, aligning directly with the terms in the original equation.

Triangle Similarity

Triangle similarity is a powerful geometric tool. It tells us that triangles are similar if their corresponding angles are equal and their corresponding sides are proportional. In the exercise, triangle similarity principles are applied to find relationships between different parts of triangles constructed in the semicircle.
To illustrate, once a semicircle is drawn, perpendiculars from points \(C\) and \(D\) meeting the semicircle create triangles \(AEC\) and \(DFA\). These triangles are similar by the Angle-Angle (AA) similarity criterion, because:

• Each has a right angle (\(\angle AEC = \angle DFA = 90°\)).
• They share another angle inscribed in the semicircle (\(\angle CAE = \angle FDA\)).

When two or more triangles are similar, the ratios of the lengths of their corresponding sides are equal. This means geometrically, triangle similarity can transform complex algebraic equations into simpler proportions. In this case, it helps in setting up the equation \(\frac{AE}{CE} = \frac{DF}{FA}\) derived from \(\frac{9}{x} = \frac{x}{5}\), giving a geometric way to solve it.

Geometric Problem Solving

Geometric problem solving often involves visualizing algebraic equations in terms of shapes and figures. This approach can simplify what may otherwise be complex algebraic manipulations. For the given problem, converting the algebraic equation into a geometric form using a semicircle introduces a relationship between triangle similarity and solving for an unknown using these geometric properties.
The step-by-step approach guides us in expressing the unknowns \(x\) as lengths in a triangle. These lengths are present due to the relationships expressed by the similar triangles formed within the semicircle. By understanding how the lengths on the semicircle's diameter correspond to terms in the equation, we can solve \(\frac{9}{x} = \frac{x}{5}\) for \(x\) by using simple geometric reasoning.
The blend of algebra and geometry here does not just provide a solution but also enhances comprehension by offering a visual grasp of the problem's nature. Using similarity, the problem reduces to a simple proportion \(CE = x\) and \(FA = 5\), followed by proving the equation to find that \(x = 3\sqrt{5}\), thus showcasing the strength of geometric problem solving.