Question

In the given square, a diagonal is drawn, and parallel line segments joining points on the adjacent sides are drawn on both sides of the diagonal. The length of the diagonal is $n\sqrt{2}$ cm. If the distance between consecutive line segments be $\frac{1}{\sqrt{2}}$ cm then the sum of the lengths of all possible line segments and the diagonal is

• A. $n(n+1)\sqrt{2}$ cm
• B. $n^2$ cm
• C. $n(n+1)$ cm
• D. $n^2\sqrt{2}$ cm

Answer 1

The correct option is D $n^2\sqrt{2}$ cm

Let consider the diagonal and on adjacent parallel line
Length of line
$PQ=RS=AC-(AR+SC)=AC-2AR(\because AR=SC)=AC-2PR(\because AR=PR)$
$=n\sqrt{2}-2.\frac{1}{\sqrt{2}}=n\sqrt{2}-\sqrt{2}=(n-1)\sqrt{2}$
Therefore sum of the length of all possible line segment and the diagonal
$=2\times [n\sqrt{2}+(n-1)\sqrt{2}+(n-2)\sqrt{2}+...]-n\sqrt{2}=2\times \sqrt{2}(n+(n-1)+(n-2)+...+1)-n\sqrt{2}$
$=2\sqrt{2}\times \frac{n(n+1)}{2}-n\sqrt{2}=n\sqrt{2}(n+1-1)=n^2\sqrt{2}cm$