Bunuel wrote:
In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?A. 3:2
B. 5:3
C. 2:1
D. 3:1
E. 4:1
The radius of the larger circle is (D+d)/2. Its area is \(\pi{(\frac{D+d}{2})^2}\).
The sum of the areas of the two smaller circles is \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}\).
Given that:
\(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\);
\(D^2+d^2=\frac{5}{8}(D+d)^2\);
\(3D^2+3d^2-10Dd=0\);
Divide by d^2: \(3(\frac{D}{d})^2+3-10\frac{D}{d}=0\);
Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).
Answer: D.
Hello. I was wondering if the following is the right approach:
Large circle area: pie (D/2 + d/2)^2
Small cricles: pie (D/2)^2 + pie(d/2)^2
(ignore pies since they cancel out)
Expand large circle using the quadratic template: (D/2)^2 + (d/2)^2 + (2)(D/2)(d/2)
Since small circles are 5/8 of the big circle. We can say that [Large circle] - [sum of two small circles area] = 1 - 5/8 or 3/8
Therefore:
Large: (D/2)^2 + (d/2)^2 + (2)(D/2)(d/2)
(subtract)
Small: (D/2)^2 + (d/2)^2
=> (2)(D/2)(d/2) = 3/8
Now solve from here onwards. Is this correct approach? Or am i forcing myself to the right answer by using an incorrect approach?
Thanks.