in a circular table cover of radius 32 cm , a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24 . Find the area of the design.

Asked by azamk4198 | 14th Jan, 2016, 08:59: AM

Expert Answer:

Draw space OL space perpendicular BC. In increment OLC angle OLC equals 1 half angle ACB equals 30 degree therefore angle LOC equals 60 degree sin 60 degree equals LC over OC LC equals fraction numerator square root of 3 over denominator 2 end fraction cross times 32 equals 16 square root of 3 cm BC equals 2 cross times LC equals 32 square root of 3 cm  Area space of space the space design space equals space Area space of space the space circle space minus space Area space of space the space increment ABC. equals πr squared minus fraction numerator square root of 3 over denominator 4 end fraction straight a squared space space space space space left square bracket straight a space is space the space length space of space the space side right square bracket equals 3.14 cross times 32 cross times 32 minus fraction numerator square root of 3 over denominator 4 end fraction cross times 32 square root of 3 cross times 32 square root of 3 equals 1024 open square brackets 3.14 minus fraction numerator 9 square root of 3 over denominator 4 end fraction close square brackets equals 256 open square brackets 12.96 minus 9 square root of 3 close square brackets  Area space of space the space design space is space 256 open square brackets 12.96 minus 9 square root of 3 close square brackets space cm squared

Answered by Vijaykumar Wani | 14th Jan, 2016, 12:05: PM