Q. 63.7( 330 Votes )

# 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 (shaded region).

Answer :

Radius of the circle "R" = 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 [ By the property of equilateral triangle inscribed in a circle]

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

By Pythagoras theorem,

AB^{2 }= AD^{2 }+ BD^{2}

⇒ AB^{2 }= 48^{2 }+ (AB/2)^{2}

⇒ AB^{2 }= 2304^{ }+ AB^{2}/4

⇒ 3/4 (AB^{2})^{ }= 2304

⇒ AB^{2 }= 3072

⇒ AB^{ }= 32√3 cm

Area of ΔABC = √3/4 × (32√3)^{2 }cm^{2 }= 768√3 cm^{2}

Area of circle = π R^{2}

= 22/7 × 32 × 32

= 22528/7 cm^{2}

Area of the design = Area of circle - Area of ΔABC

= (22528/7 - 768√3) cm^{2}

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