Question

The sides of a triangle are in A.P. and its area is $\dfrac{3}{5}$th of an equilateral triangle of the same perimeter. Prove that its sides are in the ratio \[3:5:7\].

Answer 1

Hint: Given that, the sides of a triangle are in A.P. Let the sides be \[a - d\], \[a\] and \[a + d\]. First, we find its area using Heron’s Formula. Now, its area is $\dfrac{3}{5}$th of an equilateral triangle of the same perimeter. Therefore equate their areas to find the value of a in terms of d, and thus find the sides of the triangle.

Complete step by step Answer:

Let the sides be \[a - d\], \[a\] and \[a + d\] (since the sides are in Arithmetic Progression)
Let s be semi-perimeter of the triangle.
Therefore, sum of the sides\[ = 3a = 2s\] , where s is semi-perimeter.
Hence, \[s = \dfrac{{3a}}{2}\]
Now using Heron’s Formula $ = \sqrt {s(s - a)(s - b)(s - c)} $ , where s is semi-perimeter and a, b, c are sides of triangle, we find the area of triangle.
Now, area of the triangle whose sides are in A.P. is given by
$ = \sqrt {s(s - a)(s - b)(s - c)} $
On substituting the values we get,
$ = \sqrt {\left( {\dfrac{{3a}}{2}} \right)\left( {\dfrac{{3a}}{2} - (a - d)} \right)\left( {\dfrac{{3a}}{2} - a} \right)\left( {\dfrac{{3a}}{2} - (a + d)} \right)} $
On opening brackets we get,
$ = \sqrt {\left( {\dfrac{{3a}}{2}} \right)\left( {\dfrac{{3a}}{2} - a + d} \right)\left( {\dfrac{{3a}}{2} - a} \right)\left( {\dfrac{{3a}}{2} - a - d} \right)} $
On simplification we get,
$ = \sqrt {\left( {\dfrac{{3a}}{2}} \right)\left( {\dfrac{a}{2} + d} \right)\left( {\dfrac{a}{2}} \right)\left( {\dfrac{a}{2} - d} \right)} $
On multiplying first and third term and on taking LCM of second and fourth term we get,
$ = \sqrt {\left( {\dfrac{{3{a^2}}}{4}} \right)\left( {\dfrac{{a + 2d}}{2}} \right)\left( {\dfrac{{a - 2d}}{2}} \right)} $
On further simplification we get,
$ = \sqrt {\left( {\dfrac{{3{a^2}}}{{16}}} \right)\left( {a + 2d} \right)\left( {a - 2d} \right)} $
Using \[(a + b)(a - b) = ({a^2} - {b^2})\], and taking square root of first term we get,
$ = \dfrac{{\sqrt 3 a}}{4}\sqrt {{a^2} - 4{d^2}} $
Since, perimeter of equilateral triangle \[ = \] perimeter of the given triangle\[ = 3a\]
Hence, side-length of the equilateral triangle is a
Now, area of the equilateral triangle will be $ = \dfrac{{\sqrt 3 }}{4}{a^2}$
Now , given that the area of this triangle is $\dfrac{3}{5}$th of an equilateral triangle of same perimeter.
\[ \Rightarrow \dfrac{{\sqrt 3 a}}{4}\sqrt {{a^2} - 4{d^2}} = \dfrac{3}{5} \times \dfrac{{\sqrt 3 }}{4}{a^2}\]
On simplification we get,
\[ \Rightarrow \sqrt {{a^2} - 4{d^2}} = \dfrac{{3a}}{5}\]
On squaring both the sides we get,
\[ \Rightarrow {a^2} - 4{d^2} = \dfrac{{9{a^2}}}{{25}}\]
On multiplying the equation with 25 we get,
\[ \Rightarrow 25{a^2} - 100{d^2} = 9{a^2}\]
On simplification we get,
\[ \Rightarrow 16{a^2} = 100{d^2}\]
Taking square root on both the sides we get,
\[ \Rightarrow 4a = 10d\]
On dividing by 4 we get,
$ \Rightarrow a = \dfrac{{5d}}{2}$
Therefore the sides of the given triangle are
\[a - d = \dfrac{{5d}}{2} - d = \dfrac{{3d}}{2}\],
\[a = \dfrac{{5d}}{2}\] ,
and \[a + d = \dfrac{{5d}}{2} + d = \dfrac{{7d}}{2}\]
Hence, The ratio of the sides is
$ = \dfrac{{3d}}{2}:\dfrac{{5d}}{2}:\dfrac{{7d}}{2}$
On cancelling common factors we get,
$ = 3:5:7$
Hence, if the sides of a triangle are in A.P. and its area is $\dfrac{3}{5}$th of an equilateral triangle of same perimeter, then its sides are in the ratio 3:5:7.

Note: Here as the type of the unknown triangle was not known so we use the Herons formula to calculate the area of that triangle. Heron’s formula: If the sides of a triangle are a, b and c, then the area of the triangle is given by $\vartriangle \, = \sqrt {s(s - a)(s - b)(s - c)} $, where s is the semi-perimeter of the triangle. $\left( {s = \dfrac{{a + b + c}}{2}} \right)$.