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To solve for the **area of the triangle**, we can apply the **sine area formula**:

\[
\text{Area} = \frac{1}{2} \times a \times b \times \sin(C)
\]

Where:
- \(a\) and \(b\) are two sides of the triangle,
- \(C\) is the included angle between the two sides.

---

From the diagram:
- Both sides are equal to **6 cm** (isosceles triangle),
- The angle between them is **41°**.

Now, plug in the values:

\[
\text{Area} = \frac{1}{2} \times 6 \times 6 \times \sin(41^\circ)
\]

---

First, evaluate \(\sin(41^\circ)\):

\[
\sin(41^\circ) \approx 0.6561
\]

Now substitute:

\[
\text{Area} = \frac{1}{2} \times 6 \times 6 \times 0.6561
\]

\[
\text{Area} = \frac{1}{2} \times 23.6196
\]

\[
\text{Area} = 11.8098 \, \text{cm}^2
\]

---

Rounded to the nearest **0.1 cm²**, the area is:

\[
\boxed{11.8 \, \text{cm}^2}
\]

---

Would you like any further details or explanations on this?  
Here are 5 related questions to extend your understanding:

1. How do you derive the sine area formula for any triangle?
2. What other trigonometric properties can be applied to solve triangle-related problems?
3. How do you find the height of this triangle without using trigonometry?
4. How does the area change if the angle is increased to 90°?
5. Can you use Heron's formula to find the area for this triangle?

**Tip:** Always double-check whether the triangle is isosceles, equilateral, or scalene—this can simplify calculations or suggest useful symmetry.

Work out the area of the triangle correct to the nearest 0.1 cm², where both sides are 6 cm and the angle between them is 41°.

Learn how to find the area of an isosceles triangle where both sides measure 6 cm and the included angle is 41°. This step-by-step guide uses the sine rule for calculating the area of a triangle and provides the final answer rounded to the nearest 0.1 cm². Perfect for students in Grades 8-10.

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