question_answer

Let ABC be an equilateral triangle. If the side BC is produced to the point D so that \[BC=2CD\], then \[\mathbf{A}{{\mathbf{D}}^{\mathbf{2}}}\]is equal to

A)  \[3C{{D}^{2}}\]                    

B)  \[4C{{D}^{2}}\]         

C)  \[5C{{D}^{2}}\]                    

D)  \[7C{{D}^{2}}\]

Correct Answer: D

Solution :

(d):- Let \[AB=BC=AC=x\] Then, \[BM=MC=\frac{x}{2}\] Where, AM is the median. Also   \[CD=\frac{x}{2}\]      \[\left( \therefore BC=2CD \right)\] Now, in \[\Delta AMC\], \[A{{M}^{2}}=A{{C}^{2}}-M{{C}^{2}}={{x}^{2}}-\frac{{{x}^{2}}}{4}=\frac{3{{x}^{2}}}{4}\] In, \[\Delta AMD\] \[A{{D}^{2}}=A{{M}^{2}}+M{{D}^{2}}=\frac{3{{x}^{2}}}{4}+{{x}^{2}}=\frac{7{{x}^{2}}}{4}\] \[=7\left( \frac{{{x}^{2}}}{4} \right)=7C{{D}^{2}}\]   \[\left( \therefore CD=\frac{x}{2} \right)\]