Question

Question #aa415

Answer 1

`sqrt3/2`

Explanation:

Given `DeltaABE` is equilateral, `=> angleEAB=60^@`
As `BD` is diagonal of the square, `=> angleABD=45^@`
`=> angleAFB=180-(60+45)=75^@`
Draw a line `FG`, which is perpendicular to `AB`, as shown in the figure.
`=> angleAFG=30^@, =>angleBFG=45^@`

Recall that `sin30=1/2, and sin60=sqrt3/2`
Let `AG=x => AF=2x, and FG=sqrt3x`
As `DeltaBGF` is isosceles, `=> FG=GB`
`=> AG+GB=x+sqrt3x=(1+sqrt3)x`
As `AG+GB=AB`
`=> (1+sqrt3)x=sqrt(1+sqrt3)`
`=> x=(sqrt(1+sqrt3))/(1+sqrt3)`

Area `ABF=1/2*FG*AB=1/2*sqrt3x*AB`
`=1/2*sqrt3*cancel(sqrt(1+sqrt3))/cancel(1+sqrt3)*cancelsqrt(1+sqrt3)`
`=sqrt3/2`