Question

In a regular octagon `DEFGHIJK` of side `B` each, points `EG` and `DJ` are joined to form a trapezoid `DEGJ`. What is the area of trapezoid `DEGJ`?

Answer 1

Area of trapezium `DEGJ` is `B^2(1.5+sqrt2)` or `2.9142B^2`

Explanation:

Let us consider the octagon as shown below.

In this figure, we have joined `EH,GJ,EG,DJ` and drawn perpendicular `IQ` on `GJ`. `EH` cuts `GJ` at `P`.

It is obvious that `DE=PQ=B`. Further as `DeltaPGH` and `DeltaQIJ` are right angled isosceles triangles, `PG=QJ=sqrt2/2B` i.e. `GJ=B+sqrt2/2B+sqrt2/2B=B(1+sqrt2)`.

Asall angles of a regular octagon are `135^@`, in isosceles triangle `EFG`, `m/_F=135^@` and `m/_FGE=(180^@-135^@)/2=22.5^@`

and hence `m/_EGP=67.5^@`

and `EP=GPxxtan67.5^@=Bsqrt2/2xx(1+sqrt2)` See details here.

Hence area of trapezium `DEGJ` is

`1/2(B+B(1+sqrt2))xxBsqrt2/2xx(1+sqrt2)`

= `1/4B^2sqrt2(2+sqrt2)(1+sqrt2)`

= `sqrt2B^2/4(2+2sqrt2+sqrt2+2)`

= `B^2(6+4sqrt2)/4`

= `B^2(1.5+sqrt2)`

= `2.9142B^2` as `sqrt2~=1.4142`