Question

Equilateral triangle ABC has side length of 1, and squares ABDE, BCHI, CAFG lie outside the triangle. What is the area of hexagon DEFGHI?

Answer 1

`(sqrt3+3)" sq.unit"`.

Explanation:

We will use, to find the Area of `DeltaABC`, the Formula,

`"Area of "DeltaABC="1/2*AB*AC*sin/_BAC.`

Observe that `Delta^s AFE, BDI, and CHG` are all congruent, so,

they all have the same Area,

`=1/2*AF*AE*sin/_EAF,`

`=1/2*1*1*sin{360^@-(90^@+90^@+60^@)},`

`=1/2*sin120^@`,

`=1/2*sin(180^@-60^@)`,

`=1/2*sin60^@`,

`=1/2*sqrt3/2,`

`=sqrt3/4.`

Also, Area of the equilateral `DeltaABC=1/2*1*1*sin60^@=sqrt3/4`.

`"Area of each square=1"`.

Hence, The Area of the Hexagon `DEFGHI`,

`=3xxsqrt3/4+sqrt3/4+3xx1`,

`=(sqrt3+3)" sq.unit"`.

Answer 2

Area of hexagon`=`Area of `Delta ABC+`Area of three squares`+`Area of three triangles `EAF, DBI, HCG`

1. Area of `Delta ABC`
   Draw a line perpendicular from vertex `A` on side `BC`. This is altitude of the triangle `ABC`. This perpendicular also bisects `angle BAC`. As each side of equilateral triangle is `=1` and each angle `=60^@`
   Altitude `=1xxcos30^@=sqrt3/2`
   Area of `DeltaABC=1/2xx"base"xx"altitude"`
   `=1/2xx1xxsqrt3/2=sqrt3/4`

2. Area of three squares.
   Each square has side `=1` and therefore has area `=1^2=1`
   Total area of three squares`=3xx1=3`

3. Area of three `Delta`s `EAF, DBI, HCG`
   For `DeltaEAF`
   Note that in angle at `A=360^@`
   This angle is equal to four angles `=60^@+90^@+90^@+angleEAF`
   Equating both we get `angleEAF=360^@-240^@=120^@`.
   Altitude of `DeltaEAF` can be found as explained in case of `Delta ABC` above
   Altitude of `DeltaEAF=1xxcos60^@=1/2`
   Half of side `EF=1xxsin60^@=sqrt3/2`
   Base `EF=2xxsqrt3/2=sqrt3`
   Area of `DeltaEAF=1/2xxsqrt3xx1/2=sqrt3/4`
   Similarly area of other two triangles is also same.

Area of hexagon`=sqrt3/4+3+(3xxsqrt3/4)`
`=>`Area of hexagon`=3+sqrt3`