Question

The triangular side walls of a flyover has been used for advertisement. The sides of wall are 13m,14m And 15m.The advrtisment yields an earnings of Rs 2000 per m2 a year. A company hired one of its wall for 6 month's. How much rent did they pay?

Answer 1

GIVEN:-
a=13m
b=14m
c=15m
SEMI-PERIMETER OF THE TRIANGLE,s=(13+14+15)/2=21m
USING HERON'S FORMULA,
AREA=√[s(s-a)(s-b)(s-c)]
          =√[21*(21-13)*(21-14)*(21-15)]
          =√[21*8*7*6]
          =√7056
AREA =84 $m^{2}$
since,1 $m^{2}$ costs Rs.2000 per year
therefore,84$m^{2}$ will cost 2000*84,that is Rs168000 per year
therefore for 6 months,i.e.1/2 year,it will cost
                                                     =168000*$\frac{1}{2}$
                                                     =Rs.84000
THUS,THE COMPANY WILL HAVE TO PAY Rs.84000 AS RENT.

Answer 2

S= 13+14+15/2 =42/2 =21
area of triangular region=. √21×8×7×6
=84m2
cost of advertisement perm2 a year =2000
cost advertisement of 6month. = 2000×6/12
cost of advertisement 84m2for 6month=2000×6×84/12=84000