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

Question

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

Accepted Answer

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#### Step 1: Define Dimensions
Let the length of the longer side (resting on the diameter of the semicircle) be equal to $$ 2a $$. The shorter side is determined by the semicircle's geometry and is given by:
$$
\sqrt{4 - a^2}.
$$

#### Step 2: Area of Rectangle
The area of the rectangle is expressed as:
$$
A = 2a \cdot \sqrt{4 - a^2}.
$$

To maximize the area, maximize $$ A^2 $$:
$$
A^2 = 4a^2 (4 - a^2).
$$
Simplify:
$$
A^2 = 16a^2 - 4a^4.
$$

#### Step 3: Maximize Area
To find the maximum value, differentiate $$ A^2 $$ with respect to $$ a $$ and equate to zero:
$$
\frac{d(A^2)}{da} = 32a - 16a^3 = 0.
$$
Factorize:
$$
16a(2 - a^2) = 0.
$$
Thus, $$ a = \sqrt{2} $$ (since $$ a > 0 $$).

#### Step 4: Calculate Ratio
For $$ a = \sqrt{2} $$:
- Longer side = $$ 2a = 2\sqrt{2} $$
- Shorter side = $$ \sqrt{4 - (\sqrt{2})^2} = \sqrt{4 - 2} = \sqrt{2}. $$
The ratio of longer to shorter side is:
$$
\frac{2\sqrt{2}}{\sqrt{2}} = 2:1.
$$

Thus, the correct answer is **A. $$ 2:1 $$**.

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