Question

The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. The area of the triangular field is $2100\sqrt{n}{m}^2$
$n$ is

Answer 1

$Theperimeteroftriangularfield=420m.Giventhattheratiosofthesidesare6:7:8.Sumoftheratios=6+7+8=21Lengthoffirstsideofthefield=\frac{6}{21}\times 420=6\times 20=120mLengthofsecondsideofthefield=\frac{7}{21}\times 420=7\times 20=140mLengthofthirdsideofthefield=\frac{8}{21}\times 420=8\times 20=160mAccordingtoHero{n}^{\prime }sformulathearea\left(A\right)oftrianglewithsidesa,b\&cisgivenasA=\sqrt[2]{2\left[s\right(s-a\left)\right(s-b\left)\right(s-c\left)\right]}where2s=(a+b+c).Herea=120m,b=140m,c=160m,s=\frac{(120+140+160)}{2}=\frac{420}{2}=210Areaoftriangularfield=\sqrt[2]{2[210\times (210-120)(210-140)(210-160)]}=\sqrt[2]{2(210\times 90\times 70\times 50)}=\sqrt{(3\times 7\times 3\times 3\times 7\times 5\times 10000)}=\sqrt[2]{2\left[\right(7\times 3\times 100\times 7\times 3\times 100)\times 3\times 5]}=2100\sqrt[2]{215}$