9th Class
Mathematics
Triangles
Question Bank
Triangle
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, \[\Delta AMD\] \[A{{D}^{2}}=A{{M}^{2}}+M{{D}^{2}}=\frac{3{{x}^{2}}}{4}+{{x}^{2}}=\frac{7{{x}^{2}}}{4}\] \[=7\left( \frac{{{x}^{2}}}{4} \right)=7C{{D}^{2}}\] \[\left( \therefore CD=\frac{x}{2} \right)\]