Question

From a square of side 2 cm, equal triangles are cut from its corners to form a regular octagon. we will get an octagon. what is the area of that octagon?

Answer 1

Answer:

Area of octagon  = 2.63 cm²

Explanation:

Let x = side of the regular octagon. By symmetry, each figure cut from the corners must be an isosceles right (45-45-90) triangle,

and all four of them must be congruent. Then the leg of each isosceles triangle must have measure $\frac{x}{\sqrt{2} }$

Then we must have,

$\frac{x}{\sqrt{2} } + x +  \frac{x}{\sqrt{2} }$ = 2

Now simplify this,

$\sqrt{2}*x + x = 2$

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

$x =\frac{2}{1 + \sqrt{2} }$

Side of octagon =\frac{2}{1 + \sqrt{2} }   = side of the triangle

Area of each triangle  = $\frac{1}{2} x^{2}$

                                    = (\frac{2}{1 + \sqrt{2} })²/2

                                    = $\frac{2}{2\sqrt{2} +3}$

Now area of 4 triangles =   $4*\frac{2}{2\sqrt{2} +3}$ = \frac{8}{2\sqrt{2} +3}

Area of octagon = Area of square - Area of 4 triangles

                           = 4 - $\frac{8}{2\sqrt{2} +3}$ = $16\sqrt{2} -20 = 2.63$ cm²

Area of octagon  = 2.63 cm²

Answer 2

Area of octagon  = 2.63 cm²

Explanation:

Let x = side of the regular octagon. By symmetry, each figure cut from the corners must be an isosceles right (45-45-90) triangle,

and all four of them must be congruent. Then the leg of each isosceles triangle must have measure  

Then we must have,

= 2

Now simplify this,

Side of octagon =\frac{2}{1 + \sqrt{2} }   = side of the triangle

Area of each triangle  =  

                                   = (\frac{2}{1 + \sqrt{2} })²/2

                                   =  

Now area of 4 triangles =    = \frac{8}{2\sqrt{2} +3}

Area of octagon = Area of square - Area of 4 triangles

                          = 4 -  =  cm²

Area of octagon  = 2.63 cm²