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The perimeter of strangulation field is 540m and its sides are in the ratio 25;17;12. Find the area of the triangle. Also find the cost of ploughing the field at Rs.18.80 per 10m sq.....

$L e t s i d e s o f t h e t r i a n g l e a r e a, b a n d c . G i v e n, s i d e s o f t r i a n g l e a r e i n t h e r a t i o 25 : 17 : 12 S o, a : b : c = 25 : 17 : 12 \Rightarrow a = 25 x, b = 17 x a n d c = 12 x N o w, P e r i m e t e r = 540 \Rightarrow 25 x + 17 x + 12 x = 540 \Rightarrow 54 x = 540 \Rightarrow x = 10 T h u s, s i d e o f t h e t r i a n g l e a r e, a = 25 \times 10 = 250, b = 17 \times 10 = 170 c = 12 \times 10 = 120 N o w, P e r i m e t e r = 540 \Rightarrow 2 s = 540 \Rightarrow s = 270 A r e a o f t r i a n g l e = \sqrt{s (s - a) (s - b) (s - c)} \Rightarrow A r e a o f t r i a n g l e = \sqrt{270 (270 - 250) (270 - 170) (270 - 120)} \Rightarrow A r e a o f t r i a n g l e = \sqrt{270 (20) (100) (150)} \Rightarrow A r e a o f t r i a n g l e = \sqrt{3^{3} \times 10 \times 10 \times 2 \times 10^{2} \times 3 \times 5 \times 10} = \sqrt{3^{4} \times 10^{6}} = 3^{2} \times 10^{3} \Rightarrow A r e a o f t r i a n g l e = 9000 m^{2} A n d, C o s t o f p l o u g h i n g t h e f i e l d a t R s 18.80 p e r 10 m^{2} = (A r e a o f t r i a n g l e) \times (c R a t e p e r s q . m) = 9000 \times \frac{18.80}{10} = 9000 \times 1.880 = 16929 H e n c e, C o s t o f p l o u g h i n g t h e f i e l d a t R s 18.80 p e r 10 m^{2} i s R s 16929$

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