Question

The difference between the area of a square and that of an equilateral triangle on the same base is \[\dfrac{1}{4}c{m^2}\]. What is the length of the side of the triangle?
A) \[{\left( {\dfrac{{{\text{4 - }}\sqrt {\text{3}} }}{{{\text{13}}}}} \right)^{\dfrac{{\text{1}}}{{\text{2}}}}}\]cm
B) \[{\left( {\dfrac{{{\text{4 + }}\sqrt {\text{3}} }}{{{\text{13}}}}} \right)^{\dfrac{{\text{1}}}{{\text{2}}}}}\]cm
C) \[{\left( {\dfrac{{{\text{4 + }}\sqrt {\text{3}} }}{{{\text{13}}}}} \right)^{\dfrac{{{\text{ - 1}}}}{{\text{2}}}}}\]cm
D) \[{\left( {\dfrac{{{\text{4 + }}\sqrt {\text{3}} }}{{{\text{13}}}}} \right)^{\dfrac{{{\text{ - 1}}}}{{\text{2}}}}}\]cm

Answer 1

Hint: First we assume the length of the side of triangles and square. As the given triangle is equilateral triangle all three sides have the same length and in the square of all the four sides are equal so let the length of sides of both the triangle and square be a. After that use the given data and simplify it in order to get the required answer.

Complete step by step solution: Let us assume the length of side \[ = a\]
Area of a square \[ = {a^2}\]
Area of an equilateral triangle \[ = \dfrac{{\sqrt 3 }}{4}{a^2}\]
As it is given that the difference between area of a square and that of an equilateral triangle on the same base is \[\dfrac{1}{4}c{m^2}\]
\[\therefore {a^2} - \dfrac{{\sqrt 3 }}{4}{a^2} = \dfrac{1}{4}\]
On taking \[{a^2}\] we get,
\[ \Rightarrow \left( {1 - \dfrac{{\sqrt 3 }}{4}} \right){a^2} = \dfrac{1}{4}\]
On taking LCM we get,
\[ \Rightarrow \left( {\dfrac{{4 - \sqrt 3 }}{4}} \right){a^2} = \dfrac{1}{4}\]
On cross multiplication we get,
\[ \Rightarrow {a^2} = \dfrac{1}{{4 - \sqrt 3 }}\]
On rationalizing the denominator we get,
\[ \Rightarrow {a^2} = \dfrac{1}{{4 - \sqrt 3 }} \times \dfrac{{4 + \sqrt 3 }}{{4 + \sqrt 3 }}\]
On simplification we get,
\[ \Rightarrow {a^2} = \dfrac{{4 + \sqrt 3 }}{{16 - 3}}\]
\[ \Rightarrow {a^2} = \dfrac{{4 + \sqrt 3 }}{{13}}\]
On taking root we get,
\[ \Rightarrow a = {\left( {\dfrac{{4 + \sqrt 3 }}{{13}}} \right)^{\dfrac{1}{2}}}\]cm
Thus, side of a triangle is \[{\left( {\dfrac{{4 + \sqrt 3 }}{{13}}} \right)^{\dfrac{1}{2}}}\]cm

Hence, option B \[{\left( {\dfrac{{4 + \sqrt 3 }}{{13}}} \right)^{\dfrac{1}{2}}}\]cm is correct answer.

Note: when there is an irrational number in denominator we have to rationalize it by multiplying its conjugate. The following steps show the rationalization process.
\[\dfrac{1}{{a - \sqrt b }} = \dfrac{1}{{a - \sqrt b }} \times \dfrac{{a + \sqrt b }}{{a + \sqrt b }} = \dfrac{{a + \sqrt b }}{{{a^2} - {{\left( {\sqrt b } \right)}^2}}}\]
\[\dfrac{1}{{a + \sqrt b }} = \dfrac{1}{{a + \sqrt b }} \times \dfrac{{a - \sqrt b }}{{a - \sqrt b }} = \dfrac{{a - \sqrt b }}{{{a^2} - {{\left( {\sqrt b } \right)}^2}}}\]