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

Asked by Pragya Singh | 1 year ago |  92

##### Solution :-

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

= BD = $$\dfrac{AB}{2}$$

Since, AD is the median of the triangle

AO = Radius of the circle = ($$\dfrac{2}{3}$$) AD

⇒ ($$\dfrac{2}{3}$$) AD = 32 cm

By Pythagoras theorem,

⇒ AB= 482+$$( \dfrac{AB}{2})^2$$

⇒ AB= 2304+$$\dfrac{AB^2}{4}$$

$$\dfrac{3AB^2}{4}$$= 2304

⇒ AB= 3072

⇒ AB= $$32\sqrt{3}$$ cm

Area of ΔADB = $$\dfrac{\sqrt{3}}{4}$$ ×(32$$\sqrt{3}$$)cm$$768 \sqrt{3}$$cm2

Area of circle = πR2 = ($$\dfrac{22}{7}$$)×32×32 = $$\dfrac{22528}{7}$$ cm2

Area of the design = Area of circle – Area of ΔADB

= ($$\dfrac{22528}{7}$$ – $$768 \sqrt{3}$$) cm2

Answered by Abhisek | 1 year ago

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