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:
Question:
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:
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:
Options
Answer: (1) 2:1
Explanation:
Semicircle radius = 2. Put a rectangle with its base on the diameter: let half-base = x, height = y. Point on semicircle satisfies y = sqrt(R^2 − x^2) with R = 2. Area A = full base × height = (2x)·y = 2x·sqrt(R^2 − x^2). Maximize A: consider derivative or set d/dx of x^2(R^2 − x^2) = 0. This gives x^2 = R^2/2 → x = R/√2. So base = 2x = 2R/√2 = √2·R, and height = y = R/√2. Ratio (largest side)/(smallest side) = base / height = (√2·R) / (R/√2) = 2.
Explanation:
Semicircle radius = 2. Put a rectangle with its base on the diameter: let half-base = x, height = y. Point on semicircle satisfies y = sqrt(R^2 − x^2) with R = 2. Area A = full base × height = (2x)·y = 2x·sqrt(R^2 − x^2). Maximize A: consider derivative or set d/dx of x^2(R^2 − x^2) = 0. This gives x^2 = R^2/2 → x = R/√2. So base = 2x = 2R/√2 = √2·R, and height = y = R/√2. Ratio (largest side)/(smallest side) = base / height = (√2·R) / (R/√2) = 2.
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