Answer 1

The designed area is the common region between two sectors BAEC and DAFC.

Area of sector BAEC = `90^@/360^@ xx 22/7xx(8)^2`

`=1/4xx22/7xx64`

`=(22xx16)/7 cm^2`

`= 352/7 cm^2`

Area of ΔBAC = `1/2xxBAxxBC`

`= 1/2xx8xx7 = 32 cm^2`

Area of the designed portion = 2 × (Area of segment AEC)

= 2 × (Area of sector BAEC − Area of ΔBAC)

`= 2xx(352/7 - 32) = 2((352-224)/4)`

`= (2xx128)/7`

`= 256/7 cm^2`