Question

ABCD is a square. M is midpoint of AB and N is midpoint of BC. Join DM and AN form new sides which meet at O. If \[AB = x\] then, value of AN?
A. \[AN = \dfrac{{\sqrt 5 }x}{4}cm\]
B. \[AN = \dfrac{{\sqrt 5 }x}{2}cm\]
C. \[AN = \dfrac{{\sqrt 3 }x}{2}cm\]
D. \[AN = \dfrac{{\sqrt 3 }x}{4}cm\]

Answer 1

Hint: In this problem, we need to apply the Pythagoras theorem in triangle ABN to obtain the value of side AN. In a right-angle triangle, the sum of square of the perpendicular and base is equal to the square of the hypotenuse.

Complete step by step solution:
Draw a square ABCD in which the point M is the midpoint of side AB and point N is the midpoint of side BC. Now, join the point DM and AN as shown below.

Now, apply Pythagoras theorem in \[\Delta ABN\] to obtain the value of AN.
\[
  \,\,\,\,\,\,A{N^2} = A{B^2} + B{N^2} \\
   \Rightarrow AN = \sqrt {A{B^2} + B{N^2}} \\
   \Rightarrow AN = \sqrt {{x^2} + {{\left( {\dfrac{x}{2}} \right)}^2}} \\
   \Rightarrow AN = \sqrt {{x^2} + \dfrac{{{x^2}}}{4}} \\
   \Rightarrow AN = \sqrt {\dfrac{{5{x^2}}}{4}} \\
   \Rightarrow AN = \dfrac{{\sqrt 5 x}}{2} \\
\]

Thus, the length of the side AN is \[\dfrac{{\sqrt 5 x}}{2}\], hence, option (B) is the correct answer.

Note: Pythagoras theorem is applied in a right-angle triangle. In this problem, the \[\Delta ABN\] is a right-angle triangle. The hypotenuse is the longest side in a right-angle triangle.