BMAT 2012 Section 2 Question 4

The design in the diagram is formed from four circles which all touch at the top of the shape (the diagram is not to scale).The diameter of the smallest circle is d, the second is 2d, the third is 3d and the largest is 4d. Find an expression for the area of the shading in the design.

Answer:

4) A


Radius of largest circle = 4d ÷ 2 = 2dArea of largest circle = π (2d)² = π 4d²

Radius of third largest circle = 3d ÷ 2 = 1½d
Area of third largest circle = π (1½ d)² = π 2¼d²

Radius of second largest circle = 2d ÷ 2 = 1d = d
Area of second largest circle = π (d)² = π d²

Radius of smallest circle = d ÷ 2 = ½d
Area of smallest circle = π (½d)² = π ¼d²

Area of the larger shaded area = π 4d² – π 2¼d² = πd² (4 – 2¼) = π 1¾ d²

Area of the smaller shaded area =  π d² – π ¼d² = πd² (1 – ¼) =  π ¾ d²

Total shaded area = π 1¾ d² + π ¾ d² = πd² (1¾ + ¾) = πd² 2½ = ⁵⁄₂ πd²  (2½ = ⁵⁄₂)

Sami Qamar

I’m Sami Qamar. I’m a YouTuber, Blogger, and first year med student.

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