Bunuel wrote:
In the adjoining figure there are three semicircles in which BC = 6 cm and \(BD = 6 \sqrt{3}\) cm. What is the area of the shaded region (in cm) ?
A. 9π
B. 12π
C. 15π
D. 27π
E. 28π
Attachment:
The attachment 2023-01-05_13-02-57.png is no longer available
\(\triangle ADC \) is a right angled triangle as AC is the diameter of the semicircle.
In \(\triangle ADC \)
\(BD^2 = AB * BC\)
We can prove \(BD^2 = AB * BC\) using Pythagoras theorem
- \(AB^2 + BD^2 = AD^2\)
- \(BC^2 + BD^2 = CD^2\)
- \(AC^2 = AD^2 + DC^2\)
Using the third equation -
\((AB + BC)^2 = AD^2 + DC^2\)
\(AB^2 + BC^2 + 2 AB. BC = AD^2 + DC^2\) ---- (4)
Adding equation 1 and 2
\(AB^2 + BC^2 + 2 BD^2 = AD^2 + DC^2\)
\(AB^2 + BC^2 = AD^2 + DC^2 - 2 BD^2\) ---- (5)
Substitute the value of \(AB^2 + BC^2\) from eq. (5) in eq. (4)
\(AD^2 + DC^2 - 2 BD^2 + 2 AB . BC = AD^2 + DC^2\)
Subtract \(AD^2 + DC^2\) from both sides and then divide the both sides of the equation by 2
\(BD^2 = AB. BC\)
Hence AB = 18 cms
- Diameter of the larger semi-circle (indicated by 1) = 18 + 6 = 24
Radius of the this circle (indicated by 1) = 12
- Diameter of one of the smaller semi-circle (indicated by 2) = 18
Radius of the this circle (indicated by 2) = 9
- Diameter of one of the smaller semi-circle (indicated by 3) = 6
Radius of the this circle (indicated by 3) = 3
Area of the shaded region = Area of the semi-circle 1 - (Area of semi-circle 2 + Area of semi-circle 3)
\(= 27\pi\)
The working is shown in the attachment.
Option D
Attachments
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