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**Soroban** Hello, magentarita!

Let $\displaystyle x$ = side of the triangle.

The area of an equilateral triangle of side $\displaystyle x$ is: .$\displaystyle A_t \:=\:\frac{\sqrt{3}}{4}x^2$

The triangle uses $\displaystyle 3x$ of the wire.

This leaves $\displaystyle 10-3x$ for the circumference of the circle.

Circumference formula: .$\displaystyle C \:=\:2\pi r$

So we have: .$\displaystyle 10-3x \:=\:2\pi r \quad\Rightarrow\quad r \:=\:\frac{10-3x}{2\pi} $

Area formula: .$\displaystyle A \:=\:\pi r^2$

So we have: .$\displaystyle A_c \;=\;\pi\left(\frac{10-3x}{2\pi}\right)^2 \quad\Rightarrow\quad A_c\;=\;\frac{(10-3x)^2}{4\pi}$

Therefore: . $\displaystyle \boxed{A \;=\;\frac{\sqrt{3}}{4}x^2 + \frac{(10-3x)^2}{4\pi}} $

Domain: .$\displaystyle 0 \,\leq\,x\,\leq\,3\tfrac{1}{3}$