Question

sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540m. Find its area

Answer 1

Let sides of triangle be 12n, 17n, 25n.
Perimeter of triangle is (12n + 17n + 25n) = 540  or 54n = 540             ==> n = 10. Hence sides of triangle are 120, 170, 250.
Area of triangle can be found by using formula Area = √[s(s-a)(s-b)(s-c)] where a, b, c are sides of the triangle and s = (a+b+c)/2. 
Here s = (12+17+25)/2 = 54/2 = 27.    
s-a = 27-12 = 15, s-b = 27-17 = 10, s-c = 27-25 = 2.
Hence Area = √(27x15x10x2) = √[(3x3x3)(3x5)(5x2)x2] 
                                              = √[(3x3)x(3x3)(5x5)(2x2)
                                              = 3 x 3 x 5 x2 = 90 sqm

Answer 2

Sides are in the ratio: $a:b:c = 12:17:25 \\ So\ let\ us\ say\ \\ \\ a = \frac{12x}{12+17+25} = \frac{12 x}{ 54} \\ \\ b = \frac{17 x}{54} and\ \ \ c = \frac{25 x}{ 54} \\ \\ \\ Now\ perimeter = a+b+c = \frac{(12+17+25 ) x}{ 54} = x = 540 meters \\ \\ Sides\ are:\ a = \frac{12 * 540}{54} = 120 meters \\ \\ b = 170 meters\ and\ c\ = 250 meters \\ \\ semi-perimeter\ s = (12+17+25)/2 = 27 meters$

$Area = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27*(27-12)(27-17)(27-25) } \\ \\ = \sqrt{27*15*10*2} = \sqrt{9*3*3*5*5*2*2} = 3*3*5*2 = 90 meter^2$