The ratio of the area of the inscribed circle to the area of the circumscribed circle of an equilateral triangle is____

The correct answer is 1/4.

Step-by-step Explanation:

Let us analyze the geometry of the equilateral triangle and its circles based on the provided diagrams:
1. Image 1 shows an equilateral triangle with a circle inscribed inside it (incircle) and another circle circumscribed around it (circumcircle). Both circles share the same center point, labeled as $O$ in Image 2.
2. In Image 2, the side at the bottom is labeled with vertices $A$ and $C$, and the point of tangency of the inscribed circle on $AC$ is marked as $B$.
3. The radius of the inscribed circle is represented by $r=OB$.
4. The radius of the circumscribed circle is represented by $R=OA$.
5. Since $OB$ is perpendicular to the tangent line $AC$, the triangle $OBA$ is a right-angled triangle at vertex $B$ (angle $\&\#x2220;OBA=90\&\#x2218;$).

In an equilateral triangle, each interior angle is $60\&\#x2218;$. The line segment $OA$ bisects the angle at vertex $A$, which gives:

$\&\#x2220;OAB=\frac{60\&\#x2218;}{2}=30\&\#x2218;$

Now, using the trigonometric sine function in the right-angled triangle $OBA$:

$\sin \left(30\&\#x2218;\right)=\frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{OB}{OA}=\frac{r}{R}$

Since we know that $\sin (30\&\#x2218;)=\frac{1}{2}$, we can substitute this value:

$\frac{r}{R}=\frac{1}{2}$

This simplifies to:

$R=2r$

Next, we calculate the area of both circles:
- Area of the inscribed circle:

${\mathrm{Area}}_{\text{inscribed}}=\mathrm{\&\#x3C0;}{r}^2$

- Area of the circumscribed circle:

${\mathrm{Area}}_{\text{circumscribed}}=\mathrm{\&\#x3C0;}{R}^2=\mathrm{\&\#x3C0;}{\left(2r\right)}^2=4\mathrm{\&\#x3C0;}{r}^2$

Finally, we find the ratio of the area of the inscribed circle to the area of the circumscribed circle:

$\text{Ratio}=\frac{{\mathrm{Area}}_{\text{inscribed}}}{{\mathrm{Area}}_{\text{circumscribed}}}=\frac{\mathrm{\&\#x3C0;}{r}^2}{4\mathrm{\&\#x3C0;}{r}^2}=\frac{1}{4}$