Context: In a triangle, one of the sides is 6 cm and the other two sides are in the ratio of 1: 2. Find the maximum area of the triangle.
Answer question

Let's find the maximum area of a triangle given that one of its sides is 6 cm and the other two sides are in the ratio of 1:2.

### Step 1: Express the sides in terms of a variable
Let the two sides in the ratio \(1:2\) be \(x\) and \(2x\).

Thus, the sides of the triangle are \(6 \, \text{cm}\), \(x\), and \(2x\).

### Step 2: Use the formula for the area of a triangle
The area \(A\) of a triangle with sides \(a\), \(b\), and \(c\) can be found using Heron's formula:

\[
s = \frac{a + b + c}{2}
\]

\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]

where \(s\) is the semi-perimeter of the triangle.

### Step 3: Calculate the semi-perimeter
In our case:

\[
s = \frac{6 + x + 2x}{2} = \frac{6 + 3x}{2} = \frac{3x + 6}{2}
\]

### Step 4: Substitute into Heron's formula
The area \(A\) is:

\[
A = \sqrt{\frac{3x + 6}{2} \left(\frac{3x + 6}{2} - 6\right)\left(\frac{3x + 6}{2} - x\right)\left(\frac{3x + 6}{2} - 2x\right)}
\]

Simplify each term:

- \(\frac{3x + 6}{2} - 6 = \frac{3x + 6 - 12}{2} = \frac{3x - 6}{2} = \frac{3(x - 2)}{2}\)
- \(\frac{3x + 6}{2} - x = \frac{3x + 6 - 2x}{2} = \frac{x + 6}{2}\)
- \(\frac{3x + 6}{2} - 2x = \frac{3x + 6 - 4x}{2} = \frac{-x + 6}{2} = \frac{6 - x}{2}\)

Thus, the area formula becomes:

\[
A = \sqrt{\frac{3x + 6}{2} \times \frac{3(x-2)}{2} \times \frac{x+6}{2} \times \frac{6-x}{2}}
\]

### Step 5: Find the maximum area
To find the maximum area, it's easier to use trigonometric methods instead of Heron's formula directly. Consider the area of a triangle with sides \(a\), \(b\), and \(c\) as:

\[
A = \frac{1}{2} ab \sin C
\]

For maximum area, \(\sin C = 1\), implying \(C = 90^\circ\). This means the triangle is a right triangle.

Let's assume the 6 cm side is the hypotenuse:

\[
\text{Hypotenuse} = 6, \quad \text{Legs} = x \, \text{and} \, 2x
\]

By the Pythagorean theorem:

\[
x^2 + (2x)^2 = 6^2
\]

\[
x^2 + 4x^2 = 36
\]

\[
5x^2 = 36
\]

\[
x^2 = \frac{36}{5} = 7.2
\]

\[
x = \sqrt{7.2} \approx 2.68 \, \text{cm}
\]

The other leg is \(2x = 2 \times 2.68 = 5.36 \, \text{cm}\).

### Step 6: Calculate the maximum area
The maximum area is when the triangle is right-angled:

\[
A_{\text{max}} = \frac{1}{2} \times 2.68 \times 5.36 = \frac{1}{2} \times 14.3648 = 7.18 \, \text{cm}^2
\]

### Final Answer:
The maximum area of the triangle is approximately **7.18 cm²**.

Would you like more details or have any questions?

**Relative Questions:**

1. What would the area be if the sides were in a different ratio?
2. How does the Pythagorean theorem help in finding the maximum area?
3. What happens to the area if the triangle is not a right triangle?
4. Can the 6 cm side be anything other than the hypotenuse?
5. How would the solution change if one side was specified as the base instead?

**Tip:** In optimization problems involving geometry, consider special cases like right angles or symmetry to simplify calculations.

Learn how to maximize the area of a triangle with sides in a 1:2 ratio and one side measuring 6 cm using geometric and trigonometric methods. This problem explores Heron's formula, the Pythagorean theorem, and conditions for maximizing triangle area. Suitable for high school students in Grades 10-12.

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