Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions. She divided the field in two par

Explanation

Let ABC be the triangular field where wheat is grown . Let ACD be the adjacent field which has been divided in two parts by joining C to the mid-point of AD.
Since CE is a median of `DeltaACD`. Therefore, it divides `DeltaACd` in two triangles of equal area i.e. Area of `DeltaACE=`Area of `DeltaCDE`.
Formula
Computation of the area of `DeltaABC` :
Let's be the semi-perimeter of `DeltaABC`.
Then, `s=(AB+BC+CA)/(2)=(360+200+240)/(2)m=400 m`
`:. "Area of " DeltaABC=sqrt(400xx(400-360)xx(400-200)xx(400-240))m^(2)`
`rArr "Area of" DeltaABC=sqrt(400xx40xx200xx160)m^(2)`
`rArr "Area of " DeltaABC=16000sqrt(2)m^(2)=22627.41 m^(2)=(22627.41)/(10000)`hectares
`Area "Area of "DeltaABC=2.26` hectares
Computation of the area of `DeltaACD`:
Let's be the semi-perimeter of `DeltaACD`.
Then, `s'=(Ac+CD+AD)/(2)=(240+320+400)/(2)m=480m`
`:. "Area of "DeltaACD=sqrt(480xx(480-320)xx(480-400)xx(480-240))m^(2)`
`Delta "Area of "DeltaACD=sqrt(480xx160xx80xx240)m^(2)=38400 m^(2)`
`rArr "Area of" DeltaACD=(38400)/(10000)` hectares = 3.84 hectares
So, area of `DeltaCDE=(3.84)/(2)` hectares = 1.92 hectares
Hence, Area used for growing wheat = Area of `DeltaABC=2.26` hectares
Area used for growing potatoes = Area of `DeltaACE =1.92` hectares
Area used for growing onions = Area of `DeltaCDE=1.92` hectares