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The sides of a triangle are in A.P. and its area is \(\frac{3}{5}\)th the area of an equilateral triangle of the same perimeter. The sides of the triangle are in the ratio.

(a) 1 : 2 :√7 

(b) 2 : 3 : 5 

(c) 1 : 6 : 7 

(d) 3 : 5 : 7

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Answer : (d) 3 : 5 : 7

Let the sides of the triangle be a – d, a, a + d.   

Perimeter of the triangle = a – d + a + a + d = 3a

∴ Each side of the equilateral triangle = \(\frac{3a}{3}\) = a 

∴ Area of equilateral triangle = \(\frac{\sqrt{3}}{4}a^2\) 

Area of the given ∆ = \(\sqrt{s(s-(a-d)(s-a)(s-(a+d))}\) where s = \(\frac{3a}{2}\) 

Given, \(\sqrt{\frac{3}{4}a^2(\frac{1}{4}a^2 -d^2)}\) = \(\frac{3}{5} \times\) \(\frac{\sqrt{3}}{4}a^2\) 

⇒ \(\frac{25a^2 - 9a^2}{400}\) = \(\frac{1}{4}d^2\) 

⇒ \(\frac{16a^2}{400}\) = \(\frac{1}{4}d^2\) 

⇒ \(\frac{a^2}{d^2}\) = \(\frac{400}{64}\) 

⇒ \(\frac{a}{d}\) = \(\frac{20}{8}\) = k (say)

∴ (a – d) : a : (a + d) = (20k – 8k) : 20k : (20k + 8k) 

= 12k : 20k : 28k 

= 3 : 5 : 7.

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