Thank you
-
you have a 20cm side so by pythagorean it is equal to c and the 2 other sides are equal so let it be x...
so 20^2=2x^2
400/2=x^2
sqrt 200=x =14.1421356cm
now the value of x is the 2 legs of 4 corner triangles of the big square, the area is
At=1/2*200cm^2 =100cm^2 each,so 4 is
At=400cm^2... then
multiply x by 2 and add 20cm then square it so that you get the area of big square:
As=[2x plus 20]^2
As=[(2*14.1421356) plus 20]^2 ==>
As=48.2842712^2= 2331.3708453cm^2 or 2331.371cm^2
then subtract the area of big square to area of 4 small triangles and that will be the area of octagon:
Ao=As-At
Ao=2331.371cm^2-400cm^2
Ao=1931.371cm^2
you have 2 octagon (above/below) so 2Ao=3862.742cm^2
then face area
Af=8*20*27
Af=4320cm^2
total surface area
SA=Af plus 2Ao
SA=4320cm^2 plus 3862.742 cm^2
SA=8182.742cm^2 answer (@_@)
Then volume
V=Ao*h
V=1931.371cm^2*27cm
V=52,147.017cm^3 answer (@_@)
so 20^2=2x^2
400/2=x^2
sqrt 200=x =14.1421356cm
now the value of x is the 2 legs of 4 corner triangles of the big square, the area is
At=1/2*200cm^2 =100cm^2 each,so 4 is
At=400cm^2... then
multiply x by 2 and add 20cm then square it so that you get the area of big square:
As=[2x plus 20]^2
As=[(2*14.1421356) plus 20]^2 ==>
As=48.2842712^2= 2331.3708453cm^2 or 2331.371cm^2
then subtract the area of big square to area of 4 small triangles and that will be the area of octagon:
Ao=As-At
Ao=2331.371cm^2-400cm^2
Ao=1931.371cm^2
you have 2 octagon (above/below) so 2Ao=3862.742cm^2
then face area
Af=8*20*27
Af=4320cm^2
total surface area
SA=Af plus 2Ao
SA=4320cm^2 plus 3862.742 cm^2
SA=8182.742cm^2 answer (@_@)
Then volume
V=Ao*h
V=1931.371cm^2*27cm
V=52,147.017cm^3 answer (@_@)