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The problem presents an equilateral triangle with side length \( x \) cm and an area of 36 cm². We are tasked with finding the value of \( x \), correct to 3 significant figures.

### Step 1: Formula for the Area of an Equilateral Triangle
The formula for the area \( A \) of an equilateral triangle with side length \( s \) is:

\[
A = \frac{s^2 \sqrt{3}}{4}
\]

Here, \( s = x \), and the area is given as 36 cm². So, we can set up the equation:

\[
36 = \frac{x^2 \sqrt{3}}{4}
\]

### Step 2: Solve for \( x \)
Multiply both sides of the equation by 4 to eliminate the denominator:

\[
144 = x^2 \sqrt{3}
\]

Next, divide both sides by \( \sqrt{3} \):

\[
x^2 = \frac{144}{\sqrt{3}}
\]

To simplify further:

\[
x^2 = \frac{144 \sqrt{3}}{3}
\]

\[
x^2 = 48 \sqrt{3}
\]

### Step 3: Compute the Value of \( x \)
Taking the square root of both sides:

\[
x = \sqrt{48 \sqrt{3}}
\]

This value can be computed more accurately with a calculator. Let me compute it for you.The value of \( x \), correct to 3 significant figures, is \( 9.12 \) cm.

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How is the formula for the area of an equilateral triangle derived?
2. What other geometric properties can be calculated for an equilateral triangle given its side length?
3. How would you calculate the height of this triangle?
4. How can the area formula change for other types of triangles?
5. Can this method be applied to solve for \( x \) in similar polygons with more sides?

**Tip:** When solving problems involving square roots, ensure to handle rounding properly to maintain precision at required significant figures.

The diagram shows an equilateral triangle with side lengths of x cm. The area of the equilateral triangle is 36 cm². Find the value of x, giving your answer correct to 3 significant figures.

Solve the problem of finding the side length of an equilateral triangle with an area of 36 cm². This detailed solution uses geometric and algebraic concepts suitable for students in Grades 9-11.

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