A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is

The correct answer/option is 2:1.

Let the semicircle be represented in a Cartesian coordinate system with its center at the origin (0, 0) and radius $R=2$ cm. The boundary of the semicircle is given by the equation:
$x^2+y^2=R^2=4$ for $y\ge 0$.

Let a rectangle be inscribed in this semicircle such that two of its vertices lie on the diameter along the x-axis (from $-x$ to $x$), and the other two vertices lie on the curved boundary of the semicircle at points $(-x,y)$ and $(x,y)$, where $x>0$ and $y>0$.

The dimensions of this rectangle are:
Length along the x-axis (horizontal side) = $2x$
Height along the y-axis (vertical side) = $y$

The area $A$ of the rectangle is:
$A=2xy$

Since the upper vertices lie on the circle, we have $x^2+y^2=4$, which gives:
$y=\sqrt{4-x^2}$

Substituting $y$ in terms of $x$ into the area equation:
$A=2x\sqrt{4-x^2}$

To maximize the area $A$, we can maximize its square $A^2$:
$f\left(x\right)=A^2=4x^2(4-x^2)=16x^2-4x^4$

Differentiating $f\left(x\right)$ with respect to $x$ and setting it to 0 for maximum value:
$\frac{df}{dx}=32x-16x^3=0$

Since $x>0$:
$32-16x^2=0$
$x^2=2\Rightarrow x=\sqrt{2}$

Now we calculate the corresponding value of $y$:
$y=\sqrt{4-x^2}=\sqrt{4-2}=\sqrt{2}$

Thus, the sides of the rectangle are:
One side = $2x=2\sqrt{2}$ cm
The other side = $y=\sqrt{2}$ cm

The larger side is $2\sqrt{2}$ and the smaller side is $\sqrt{2}$.

The ratio of the lengths of the largest to the smallest side is:
$\text{Ratio}=\frac{2\sqrt{2}}{\sqrt{2}}=\frac{2}{1}=2:1$