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A. \[\dfrac{{2\sqrt 3 }}{{15}}\]

B. \[\dfrac{{2\sqrt 3 }}{5}\]

C. \[\dfrac{{4\sqrt 3 }}{5}\]

D. \[\dfrac{{4\sqrt 3 }}{{15}}\]

Answer
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* Pythagoras theorem states that in a right angled triangle, the sum of square of base and square of height is equal to square of the hypotenuse.

If we have a right angled triangle, \[\vartriangle ABC\] with right angle, \[\angle B = {90^ \circ }\]

Then using the Pythagoras theorem we can write that \[A{C^2} = A{B^2} + B{C^2}\]

* Perpendicular drawn from the vertex of an equilateral triangle bisects the side of the triangle.

* The shortest distance from point \[({x_1},{y_1})\] to the line \[ax + by + c = 0\] is given by

\[D = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\]

We draw an equilateral triangle whose base is along the line \[3x + 4y = 9\] and has a vertex \[(1,2)\].

We know that an equilateral triangle has all sides of equal length, say ‘a’.

Then the perpendicular AD drawn from A to the base will bisect the base.

So, the length \[BD + DC = BC\]. Since the base is cut in equal halves

\[

DC + DC = a \\

2DC = a \\

DC = \dfrac{a}{2} \\

\]

Now we apply Pythagoras theorem to \[\vartriangle ADC\]

\[

\Rightarrow A{C^2} = A{D^2} + D{C^2} \\

\Rightarrow {a^2} = A{D^2} + {\left( {\dfrac{a}{2}} \right)^2} \\

\]

Shifting the constants to one side of the equation

\[ \Rightarrow {a^2} - \dfrac{{{a^2}}}{4} = A{D^2}\]

Taking LCM on LHS of the equation

\[

\Rightarrow \dfrac{{4{a^2} - {a^2}}}{4} = A{D^2} \\

\Rightarrow \dfrac{{3{a^2}}}{4} = A{D^2} \\

\]

Take square root on both sides of the equation

\[

\Rightarrow \sqrt {\dfrac{{3{a^2}}}{4}} = \sqrt {A{D^2}} \\

\Rightarrow \dfrac{{\sqrt 3 a}}{2} = AD \\

\]

So the length of perpendicular \[AD = \dfrac{{\sqrt 3 a}}{2}\] … (1)

We have point \[(1,2)\] and the shortest distance between the point and the line \[3x + 4y = 9\] is \[AD\].

Using the formula for shortest distance between a point \[({x_1},{y_1})\] and a line \[ax + by + c = 0\]

\[D = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\]

On comparing line with general equation of line we get \[a = 3,b = 4,c = - 9\] and \[{x_1} = 1,{x_2} = 2\]

\[ \Rightarrow D = \dfrac{{\left| {3 \times 1 + 4 \times 2 - 9} \right|}}{{\sqrt {{3^2} + {4^2}} }}\]

Squaring the terms in the denominator

\[

\Rightarrow D = \dfrac{{\left| {3 + 8 - 9} \right|}}{{\sqrt {9 + 16} }} \\

\Rightarrow D = \dfrac{{\left| {11 - 9} \right|}}{{\sqrt {25} }} \\

\Rightarrow D = \dfrac{{\left| 2 \right|}}{{\sqrt {{5^2}} }} \\

\]

Cancel square root with square value

\[ \Rightarrow D = \dfrac{2}{5}\]

Now we have the distance from vertex A to base BC

So, \[AD = \dfrac{2}{5}\] … (2)

Equating values of AD from equation (1) and (2)

\[ \Rightarrow \dfrac{{\sqrt 3 a}}{2} = \dfrac{2}{5}\]

Cross multiplying the terms

\[

\Rightarrow \sqrt 3 a \times 5 = 2 \times 2 \\

\Rightarrow 5\sqrt 3 a = 4 \\

\]

Shift all the values except a to one side

\[ \Rightarrow a = \dfrac{4}{{5\sqrt 3 }}\]

Rationalize by multiplying both numerator and denominator by \[\sqrt 3 \]

\[

\Rightarrow a = \dfrac{4}{{5\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} \\

\Rightarrow a = \dfrac{{4\sqrt 3 }}{{5 \times 3}} \\

\Rightarrow a = \dfrac{{4\sqrt 3 }}{{15}} \\

\]

So, the length of the side of the equilateral triangle is \[\dfrac{{4\sqrt 3 }}{{15}}\].