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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.Q6

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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

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  1. 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,

    Ncert solution class 10 chapter 12-17

    By Pythagoras theorem,

    AB= AD+BD2

    ⇒ AB= 482+(AB/2)2

    ⇒ AB= 2304+AB2/4

    ⇒ 3/4 (AB2)= 2304

    ⇒ AB= 3072

    ⇒ AB= 32√3 cm

    Area of ΔADB = √3/4 ×(32√3)cm= 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

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