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

We can see the reference solution uses a valid approach but jumps around concepts that might confuse students. Let's provide a cleaner, more student-friendly solution that maintains the same mathematical rigor.

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Let's place our semicircle with center at the origin and diameter along the x-axis from (-2, 0) to (2, 0).

For the rectangle with maximum area:

Length = a (horizontal dimension)

Breadth = b (vertical dimension)

The rectangle's top-right corner lies on the semicircle, so it satisfies the semicircle equation:

$\left(\frac{a}{2}\right)^2 + b^2 = 2^2 = 4$

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We want to maximize the area: Area = ab

Here's the clever part: We can rewrite ab using our constraint equation.

From the constraint: $\frac{a^2}{4} + b^2 = 4$

Notice that: $ab = 2 \cdot \frac{a}{2} \cdot b = 2\sqrt{\frac{a^2}{4} \cdot b^2}$

By AM-GM inequality, for any two positive numbers x and y:

$\frac{x + y}{2} \geq \sqrt{xy}$

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Let $x = \frac{a^2}{4}$ and $y = b^2$

Then: $\frac{\frac{a^2}{4} + b^2}{2} \geq \sqrt{\frac{a^2}{4} \cdot b^2}$

Since $\frac{a^2}{4} + b^2 = 4$, we have:

$\frac{4}{2} = 2 \geq \sqrt{\frac{a^2}{4} \cdot b^2} = \frac{ab}{2}$

Therefore: $ab \leq 4$

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For maximum area, equality must hold in AM-GM, which happens when:

$\frac{a^2}{4} = b^2$

This gives us: $a^2 = 4b^2$, so $a = 2b$

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Since $a = 2b$, we have:

Largest side = a = 2b

Smallest side = b

Therefore, the ratio of largest to smallest side = $\frac{a}{b} = \frac{2b}{b} = 2:1$

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Answer: 2:1