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.