A farmer has two triangular fields PQR and PQS in which side PQ is common as shown in the figure. PQ = 56 m, PR = 60 m, QR = 52 m, QS = 64 m and PS = 48 m. He has marked mid-points L and M on the sides PS and PR respectively. By joining LQ and MQ, he has made a field in the shape of quadrilateral PLQM. He grew wheat in the quadrilateral PLQM, potatoes in MQR and onions in LQS. How much area has been used for each crop?
Answers
Answer:
Ar△ACD=
s(s−a)(s−b)(s−c)
s=
2
a+b+c
=
2
800+480+640
s=
2
1920
=960m
ar△ACD=
960(960−800)(960−480)(960−640)
=
960×160×480×320
=
(100)
2
×2×3×16×16×3×16×2×16
=2×3×16×16×100=153600 m
2
ar△ACD=153600=15.36 m
2
Now, ar△CFD=ar△CFA=
2
1
×ar△ACD=
2
15.36
=7.68 hec
Because F is the mid point of AD and both △ have same height
Now, ar△ABC=
s(s−a)(s−b)(s−c)
s=
2
a+b+c
=
2
840+480+600
=960m
ar△ABC=
960(960−840)(960−480)(960−600)
=
960×120×480×360
=
(100)
2
×16×6×6×2×16×3×36
=6×6×16×100
6
≈141120 m
2
∴ar△ABC=141120 m
2
=14.11 hectare
Now, ar△AEC=ar△BEC=
2
1
×ar△ABC=
2
14.11
≈7.06
Because E is the mid point of AB and height of both triangles is also equal.
Now, area of potatoes field=ar△CFD=7.68 hectare
Area of wheat field=ar△ACF+ar△FCE
=7.68+7.06=14.74 hectare
Area of onion field=ar△BEC=7.06 hectare