question_answer

In an equilateral triangle ABC of side 10cm, the side BC is trisected at D. Then the length (in cm) of AD is

A) \[3\sqrt{7}\]                              

B) \[7\sqrt{3}\]

C) \[\frac{10\sqrt{7}}{3}\]   

D)        \[\frac{7\sqrt{10}}{3}\]

Correct Answer: C

Solution :

            \[AE\bot BC\] \[\therefore \]\[BE=EC=5\,\,cm\]             \[AC=10\,\,cm\]             \[AE=\sqrt{{{10}^{2}}-{{5}^{2}}}\]             \[=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}cm\]             \[DE=DC-EC\]             \[=\frac{2}{3}\times 10-5=\frac{5}{3}cm\] \[\therefore \]      \[AD=\sqrt{{{\left( \frac{5}{3} \right)}^{2}}+{{\left( 5\sqrt{3} \right)}^{2}}}\]             \[=\sqrt{\frac{25}{9}+75}=\sqrt{\frac{25+675}{9}}\]             \[=\sqrt{\frac{700}{9}}=\frac{10\sqrt{7}}{3}cm\]