Question

6. 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.
​

Answer 1

Step-by-step explanation:

Let ABC be the eq./and let o be the centre of the circle of r=32cm

Area of circle =tr^2 =(22/7x32x32)cm2

=22528/7 cm2

Draw OM_L_BC

Now, /_ BOM= 1/2x120°=60°

So, From ABOM,we have

OM/OB=cos 60°(1/2) i.e., OM= 16 cm

Also, BM/OB= cos60°(1/2)

i.e., BM= 16v3 cm

BC = 2 BM =32v3 cm

Hence, area of ABOC = 1/2 BC .OM =1/2X32V3x16

area of A ABC = 3x area of A BOC = 3x1/2x32v3x16

= 768 v3 cm^2

Area of design= area of O - area of ABC (22528/7 - 768 V3)cm^2