SOLUTION: the area of a triangular lot is 2,598.08 m^2. if the sides of the lot are in continued proportion of 3:5:7 , find the shortest side of the lot.

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Question 1178369: the area of a triangular lot is 2,598.08 m^2. if the sides of the lot are in continued proportion of 3:5:7 , find the shortest side of the lot.
Answer by amarjeeth123(570) About Me  (Show Source):
You can put this solution on YOUR website!
The sides of the lot are in the proportion of 3:5:7.
Let the sides be 3x,5x and 7x where x is a constant.
Let the sides be a=3x, b=5x and c=7x where a,b and c are the three sides.
Area of a triangle=sqrt%284%2Aa%5E2%2Ab%5E2-%28a%5E2%2Bb%5E2-c%5E2%29%5E2%29%2F4
Plugging in the values we get,
Area of the triangle=sqrt%284%2Aa%5E2%2Ab%5E2-%28a%5E2%2Bb%5E2-c%5E2%29%5E2%29%2F4=sqrt(675)*(x^2/4)
Area of the triangle=2,598.08 m^2
Equating both sides we get,
sqrt(675)*(x^2/4)=2,598.08 m^2
x^2=(2598.08*4)/sqrt(675)
x^2=400.0005
x=sqrt(400.0005)=20
The value of x is 20.
The shortest side of the lot is 3x which is 60m.
Answer=60m.