A farmer has a triangular field with sides 360 m, 200 m and 240 m, where he grows wheat. Adjacent to this field, he has another triangular field with sides 240 m, 320 m and 400 m, divided into two parts by joining the midpoint of the longest side to the opposite vertex. He grows potatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions ? (1 hectare = `10000 m^(2)`.)

Explanation

Let ABC be the field where wheat is grown. Let ACD be the field which has been divided into two parts by joining C to the midpoint E of AD.
For `Delta`ABC, we have
a = 200 m, b = 240 m and c = 360 m.
`therefore" "s=(1)/(2)(200 + 240 + 360)`m = 400 m.
`therefore" "(s-a)=(400-200)`m = 200 m,
(s - b) = (400 - 240)m = 160 m
and (s -c) = (400 - 360)m = 40 m.
Area of `Delta`ABC = `sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(400 xx 200 xx 160 xx 40)m^(2) = (16000 xx sqrt(2))m^(2)`
`=((16000 xx 1.41)/(10000)`hectares = 2.26 hectares.
For `Delta`ACD, we have
a = 240 m, b = 320 m and c = 400 m.
`therefore" "s=(1)/(2)(240 + 320 + 400)m =((1)/(2) xx 960)`m = 480 m.
`therefore" "(s-a)=(480 -240)`m = 240 m,
(s - b) = (480 - 320)m = 160 m
and (s - c) = (480 - 400)m = 80 m.
Area of `Delta`ACD = `sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(480 xx 240 xx 160 xx 80)m^(2)`
`=(240 xx 160)m^(2) = 38400 m^(2)`
`=(38400)/(10000)`hectares = 3.84 hectares.
Now, `Delta`ACE and `Delta`DCE have equal bases and the same height.
So, they are equal in area.