ABCD is a field in the shape of a Trapezium ab parallel to dc and angle abc is equal to 90 degree angle Dab is equal to 60 degree 4 sectors are formed with Centre ABC and in the radius of a circle is 17.5 M.
A) find the total area of four sectors
B) Find the area of remaining portion given that AB =7.5m and DC = 50m.
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Solution :-
Area of sector I = 60/360 × πr^2
Area of sector II = 90/360 × πr^2
III = 90/360×πr^2
IV =120 /360 × πr^2
Total area of 4 sectors =
=> 60+90+90+120 /360 × πr^2
=> 962.5m^2
Area of the field ABCD =1/2 h × ( a+b )
=>1/2 × 50 ×125
=> 25 ×125
=> 3125m^2
Area of remaining portion :-
=> 3125 -962.5
=>2162.5 m^2
==============
@GauravSaxena01
Area of sector I = 60/360 × πr^2
Area of sector II = 90/360 × πr^2
III = 90/360×πr^2
IV =120 /360 × πr^2
Total area of 4 sectors =
=> 60+90+90+120 /360 × πr^2
=> 962.5m^2
Area of the field ABCD =1/2 h × ( a+b )
=>1/2 × 50 ×125
=> 25 ×125
=> 3125m^2
Area of remaining portion :-
=> 3125 -962.5
=>2162.5 m^2
==============
@GauravSaxena01
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