Four equilateral triangles are described on the four sides of a rectangle with perimeter 12 cm. If the sum of the areas of the four triangles is $10\sqrt{3}$ cm$^2$, what is the area of the rectangle?

### Concept & Algebraic Manipulation of Perimeter and Area Translate the geometric properties of perimeters and equilateral triangle areas into an algebraic system of equations to solve for the rectangle's dimensions. ### Step-by-Step Solution * Let the length and breadth of the rectangle be $l$ and $b$. * The perimeter is given as 12 cm: $2(l + b) = 12 \Rightarrow l + b = 6$. * Squaring both sides: $(l + b)^2 = 36 \Rightarrow l^2 + b^2 + 2lb = 36$. * The equilateral triangles are formed on the sides. Two have side $l$ and two have side $b$. * Sum of their areas = $2 \cdot (\frac{\sqrt{3}}{4}l^2) + 2 \cdot (\frac{\sqrt{3}}{4}b^2) = 10\sqrt{3}$. * Simplify the equation: $\frac{\sqrt{3}}{2}(l^2 + b^2) = 10\sqrt{3}$. * Divide by $\sqrt{3}$: $\frac{1}{2}(l^2 + b^2) = 10 \Rightarrow l^2 + b^2 = 20$. * Substitute $l^2 + b^2 = 20$ into our squared perimeter equation: $20 + 2lb = 36$. * $2lb = 16 \Rightarrow lb = 8$. * The area of a rectangle is length $\times$ breadth ($lb$), which is 8 cm$^2$. ### Exam Strategy & Shortcut Use the identity $(l+b)^2 = l^2 + b^2 + 2lb$. From the perimeter, $l+b=6$, so $(l+b)^2 = 36$. From the triangle areas, $l^2+b^2 = 20$. The difference ($36 - 20 = 16$) equals $2lb$. Area ($lb$) is half of that, which is 8. ### Common Pitfall Trying to solve for $l$ and $b$ individually using the quadratic formula, which wastes time. Solving directly for the product $lb$ is much more efficient. ### Final Answer Therefore, the correct answer is **8 cm$^2**.