Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th

Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th
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(d) 3 : 1Let ABC be the triangular flower bed of side lengths a, b and c respectively. Then Area of ΔABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(rac{a+b+c}{2}\)Now according to the given condition,ΔPQR forms the park with side lengths 2a, 2b, 2c.∴ Area of ΔPQR = \(\sqrt{s'(s'-2a)(s'-2b)(s'-2c)}\)where s' =\(rac{2a+2b+2c}{2}\) = a + b + c = 2s∴ Area of ΔPQR = \(\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}\)= 4 \(\sqrt{s(s-a)(s-b)(s-c)}\)= 4. Area of ΔABC.∴ Area of path =  Area of ΔPQR - Area of ΔABC= 4 x Area of ΔABC - Area of ΔABC= 3 (Area of ΔABC)∴ Reqd. Ratio = Area of Path : Area of ΔABC = 3 : 1.
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