Question

A regular octagon is to be formed by cutting equal isosceles right triangles from the corners of a square. If the square has sides of one unit, the leg of each of the triangles has length:

• A. $\frac{2+\sqrt{2}}{3}$
• B. $\frac{2-\sqrt{2}}{2}$
• C. $\frac{1+\sqrt{2}}{2}$
• D. $\frac{1+\sqrt{2}}{3}$
• E. $\frac{2-\sqrt{2}}{3}$

Answer 1

Answer: (B)

$\frac{2-\sqrt{2}}{2}$

Let $ABCD$ be the square whose each side is $1$ unit.

Let $EAL,FBG,HCI$ and $JDK$ be the congruent isosceles right angled triangles.

Let $EFGHIJAKL$ be the regular octagon obtained.

Let $AE=AL=xunits$

Similarly, $BF=BG=xunits$

Now, $EF=AB-(AE+BF)$

$EF=1-(x+x)$

$\therefore$ $EF=1-2x$ ---- ( 1 )

In $△EAL,$

$(EL{)}^2=(AL{)}^2+(EA{)}^2$

$\Rightarrow$ $(EL{)}^2=x^2+x^2$

$\Rightarrow$ $(EL{)}^2=2x^2$

$\Rightarrow$ $EL=\sqrt{2}x$ ------ ( 2 )

As, the sides of a regular octagon are equal,

$\Rightarrow$ $1-2x=\sqrt{2}x$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $1=x(\sqrt{2}+2)$

$\Rightarrow$ $x=\frac{1}{\sqrt{2}+2}=\frac{\sqrt{2}-2}{(\sqrt{2}+2)(\sqrt{2}-2)}=\frac{\sqrt{2}-2}{2-4}$

$\therefore$ $x=\frac{2-\sqrt{2}}{2}$

$\therefore$ Length of the triangle $\frac{2-\sqrt{2}}{2}$.