# sonu and monu had adjacent triangular fields with common boundary 25m. The other 2 sides of sonu's field were 52m and 63m, while of monu's field were 114m and 101m . If the cost of fertilization is Rs 20per sq m, then find the total cost of fertilization for sonu and monu togeother.

s = (25+52+63)/2

s = 70

Area = √s(s-a)(s-b)(s-c)

= √70(70-25)(70-52)(70-63)

= √70(45)(18)(7)

= √396900

= 630

So, area of sonu's field is 630m

^{2}.

Area of monu's field,

s = (25+114+101)/2

s = 120

Area = √120(120-25)(120-114)(120-101)

= √120(95)(6)(19)

= √1299600 = 1140

So, area of monu's field = 1140m

^{2}.

Total area = area of monu's field + area of sonu's field

= 1140+630

= 1770m

^{2}

As it costs 20 rupees per m

^{2}

So, 1770*20

= 35400r

So, it'll cost 35400 rupees together.

Hope it'll help!!!!!!!!

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