from an internal point of equilateral …

Let the side of equilateral triangle be a and let the perpendicular drawn to sides be OP = 10 cm, OQ = 14 cm and OR = 6 cm.

Join OA, OB and OC.

According to question,

Area of equilateral triangle ABC = Area of {tex}\Delta{/tex}OBC + Area of {tex}\Delta{/tex}OCA + Area of {tex}\Delta{/tex}OAB

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2} \times {\rm{BC}} \times {\rm{OP}} + {1 \over 2} \times {\rm{AC}} \times {\rm{OQ}} + {1 \over 2} \times {\rm{AB}} \times {\rm{OR}}{/tex}

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2} \times 10 \times a + {1 \over 2} \times 14 \times a + {1 \over 2} \times 6 \times a{/tex}

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2}a\left( {10 + 14 + 6} \right){/tex}

=>     {tex}{{\sqrt 3 } \over 4}a = {1 \over 2} \times 30{/tex}

=>     {tex}a = {{15 \times 4} \over {\sqrt 3 }} = {{60} \over {\sqrt 3 }}{/tex} cm

Now, Area of equilateral triangle

{tex}{{\sqrt 3 } \over 4}{a^2} = {{\sqrt 3 } \over 4} \times {{60} \over {\sqrt 3 }} \times {{60} \over {\sqrt 3 }}{/tex}

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = 300\sqrt 3 {/tex} cm²