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

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Answer 1 · 2017-12-17T16:48:37

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

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 · 2017-12-17T17:07:13

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

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 $DeltaABC=1/2xx"base"xx"altitude"$
$=1/2xx1xxsqrt3/2=sqrt3/4$
Area of three squares.
Each square has side $=1$ and therefore has area $=1^2=1$
Total area of three squares $=3xx1=3$
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 $=3+sqrt3$