Question

Please solve q 108 ?

Answer 1

Area of an equilateral triangle `A=sqrt{3}/4 s^2`

Area ABC `= {64 sqrt{3}}/4`

Area DEF `= {9 sqrt{3} }/4`

Area ABD`= 1/3 ( 64 - 9) sqrt{3}/4 = 55/12 sqrt{3}`

Let G be the foot of the altitude on AB to D, `h=`GD

`1/2 (8) h = 55/12 sqrt{3}`

`h = 55/48 sqrt{3}`

Now we have two right triangles. Unknowns `x=`BF and `y=`BD

`x^2 + h^2 = y^2,`

`(8-x)^2 + h^2 = (y+ 3)^2`

Gotta go, gonna post; maybe continue later.

Answer 2

I get `AE+BD+CF~~9.1cm`

Explanation:

By symmetry of the figure we have `AE=BD=CF`. Let `AE=BD=CF=x` cm.

Now `DeltaABCandDeltaDEF` are equilateral having lengths of each side 8cm and 3cm respectively. Each internal angle of them is `60^@`. So for `DeltaBFC` we we have
`angleBFC=120^@,BC=8cm,CF=xcm andBF=x+3cm`

So by cosine law

`BC^2=BF^2+CF^2-2BFxxCFcosangleBFC`

`=>8^2=(x+3)^2+x^2-2(x+3)xcos120^@`

`=>8^2=x^2+6x+9+x^2-2(x+3)x(-1/2)`

`=>64=x^2+6x+9+x^2+x^2+3x`

`=>3x^2+9x-55=0`

`=>x=(-9+sqrt(9^2-4*3*(-55)))/(2*3)`
(Negative root neglected)

So

`AE+BD+CF=3x~~9.1` cm