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 parts by joining the mid point of the longest side to the opposite vertex and grew potatoes in one part and onions in one part. How much area (in hectares) has been used for wh)

Explanation

E is the midpoint ofAD,`AE=ED=200m`
Area of wheat field`=ar/_\ABC`
Area of potato field`=ar/_\AEC`
Area of onion field`=ar/_\EDC`
ar`/_\ABC= sqrt(S(S−a)(S−b)(S−c))`
`S=1/2(200+240+360)=400m`
`=sqrt(400(400−200)(400−240)(400−360))`
`=sqrt(400×200×160×40)`
`=16000sqrtm^2=1.6sqrt2hectare`
`=1.6×1.414=2.26hectares`
In `/_\ACD ,/_\AEC \and\ /_\CED` have same base `(ar AE=ED=200m)`
and they have same height `CF`
Therefore,ar`/_\AEC=ar/_\CED=1/2ar/_\ACD`
ar`/_\ACD=sqrt(S(S−a)(S−b)(S−c))`
`S=(240+320+400)/2=480m`
`= sqrt(480(480−240)(480−320)(480−400))`
`=sqrt(480×240×160×80)`
`=38400m^2=3.84hectares`
`ar /_\AEC=ar/_\CED=1/2xx3.84=1.92hectares`
Therefore,Area of wheat field`=2.26hectares`
Area of potato field`=1.92hectares`
Area of onion field`=1.92hectares`
For potato `100m=100×1.92=192hectares`