Question
The sides of a triangle are in A.P. and its area is $$\frac{3}{5} \times $$ (area of an equilateral triangle of the same perimeter). Then the ratio of the sides is
A.
1 : 2 : 3
B.
3 : 5 : 7
C.
1 : 3 : 5
D.
None of these
Answer :
3 : 5 : 7
Solution :
Here, $$2b = a + c$$ and $$\vartriangle = \frac{3}{5} \times \frac{{\sqrt 3 }}{4} \cdot {\left( {\frac{{a + b + c}}{3}} \right)^2}\,\,\,\therefore \,\,\vartriangle = \frac{{3\sqrt 3 }}{{20}} \cdot {b^2}.$$
$$\eqalign{
& \therefore \,\,\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} = \frac{{3\sqrt 3 }}{{20}}{b^2} \cr
& {\text{or, }}\sqrt {\frac{1}{{16}}\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)} = \frac{{3\sqrt 3 }}{{20}}{b^2} \cr
& {\text{or, }}\sqrt {3b \cdot \left( {b + c + c - 2b} \right)b\left( {2b - c + b - c} \right)} = \frac{{3\sqrt 3 }}{5}{b^2} \cr
& {\text{or, }}\sqrt {\left( {2c - b} \right)\left( {3b - 2c} \right)} = \frac{3}{5}b\,\,\,{\text{or, }}8bc - 4{c^2} - 3{b^2} = \frac{9}{{25}}{b^2} \cr
& {\text{or, }}\frac{{84}}{{25}}{b^2} - 8bc + 4{c^2} = 0\,\,\,\,\,{\text{or, }}\frac{b}{c} = \frac{{8 \pm \sqrt {64 - 16 \cdot \frac{{84}}{{25}}} }}{{2 \cdot \frac{{84}}{{25}}}} = \frac{5}{7},\frac{5}{3}. \cr
& {\text{Also, }}2b = a + c \cr
& \Rightarrow \,\,2\frac{b}{c} = \frac{a}{c} + 1 \cr
& \Rightarrow \,\,\frac{a}{c} = \frac{{2b}}{c} - 1 = \frac{3}{7},\frac{7}{3}. \cr} $$