Multiple choice

A circular disc of area $\displaystyle A_{1}$ is given ;with its radius as the diameter a circular disc of area $\displaystyle A_{2}$ is cut out.If the area of the remaining disc is denoted by $\displaystyle A_{3}$ Then

  1. $\displaystyle A_{1}A_{3}<16 A_{2}^{2}$
  2. $\displaystyle A_{1}A_{3}>16 A_{2}^{2}$
  3. $\displaystyle A_{1}A_{3}=16 A_{2}^{2}$
  4. $\displaystyle A_{1}A_{3}>2 A_{2}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A1 = pi*R^2. Radius of A2 is R/2, so A2 = pi*(R/2)^2 = pi*R^2/4 = A1/4. A3 = A1 - A2 = A1 - A1/4 = 3*A1/4. A1*A3 = A1*(3*A1/4) = 3*A1^2/4. 16*A2^2 = 16*(A1/4)^2 = 16*A1^2/16 = A1^2. Since 3/4 < 1, A1*A3 < 16*A2^2.

AI explanation

Let the radius of the original disc be r, so its area A1 is pi * r^2. The diameter of this disc equals 2r, making it the radius of the cut-out disc, so the area A2 is pi * (2r)^2 which equals 4pi * r^2. The area of the remaining disc is A3, calculated as A1 minus A2, yielding pi * r^2 - 4pi * r^2 = -3pi * r^2, which implies a logical error in the problem phrasing where the cut-out radius must actually be r/2. Using r/2 for the cut-out radius gives A2 = (pi * r^2)/4 and A3 = (3 * pi * r^2)/4. Multiplying A1 and A3 gives (pi * r^2) * (3 * pi * r^2 / 4) which equals 3pi^2 * r^4 / 4. The expression 16A2^2 evaluates to 16 * (pi * r^2 / 4)^2, which equals pi^2 * r^4. Comparing A1A3 to 16A2^2 shows 3/4 is less than 1, so A1A3 < 16A2^2.

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