What is the solution of ncert class 10 chapter areas related to circles , how i solve this question in a easiest and simplest way . Find the best solution of exercise 12.3 question no. 6 , sir please help me to solve this question in a easiest way.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
Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
∴ AO = Radius of the circle = (2/3) AD
⇒ (2/3)AD = 32 cm
⇒ AD = 48 cm
In ΔADB,
By Pythagoras theorem,
AB2 = AD2 +BD2
⇒ AB2 = 482+(AB/2)2
⇒ AB2 = 2304+AB2/4
⇒ 3/4 (AB2)= 2304
⇒ AB2 = 3072
⇒ AB= 32√3 cm
Area of ΔADB = √3/4 ×(32√3)2 cm2 = 768√3 cm2
Area of circle = πR2 = (22/7)×32×32 = 22528/7 cm2
Area of the design = Area of circle – Area of ΔADB
= (22528/7 – 768√3) cm2