ABCD is a field in the shape of a trapezium AB ∥ DC and ∠ABC = 90°, ∠DAB = 60°. Four sectors are formed with centers A, B, C, and D. The radius of each sector is 17.5 m. Find the area of the remaining portion, given that AB= 75m and CD= 50m.
Answers
Answered by
1
Answer:
2162.5m^2
Step-by-step explanation:
area if sector I=60/360π^2
" " " ll=90/360π^2
III = 90/360π^2
IV =120/360π^2
total area=60+90+90+120/360π^2
=962.5m^2
area of field ABCD=1/2hx(a+b)
1/2x50x125
25x125
3125m^2
area of remaining portion
3125-962.5
=2162.5 m ^2
hope it helps
please mark me brainalist
Similar questions