1)An equilateral triangle has an

1)

The  area of an equilateral triangle   is  given by

√3/4 side^2

So

64√3  =  √3 /4  side^2

256 = side^2

16 cm  = side  of original equilateral triangle

Decreasing each side by 4 cm  produces an area of  

√3/ 4 * 12^2   =  

36√3  cm^2

So  the area decrease  is    64√3  - 36√3   =  28√3 cm^2

2.

We can find angle BAC as follows

6^2  = 2(5)^2  - 2(5^2) cos(BAC)

[36 - 50]  / (-50)  = cos(BAC)

-14/-50  = cos(BAC)

7/25 = cos (BAC)

arccos ( 7/25) = BAC

And since BAC is an inscribed angle, the BOC has twice its measure =  2arccos(7/25)

And   cos [ 2 arccos (7/25) ]  =    - 527/ 625

And OC  = OB

So......using the Law of Cosines

6^2  = 2(OC)^2 - 2(OC)^2 cos [2arccos(7/25) ]

36 = 2(OC)^2 - 2(OC)^2 [ -527/ 625]

36 = 2(OC)^2 + 2(OC)^2 [ 527/625]

18 = OC^2 [ 1 + 527/625]

18 = OC^2 [ 1152/ 625]

18 * 625 / 1152  = OC^2  =  11250/ 1152  =  625/ 64

So.....the area of triangle OBC  =

(1/2) OC^2 * sin (BOC)

(1/2) (625/64) * sin [ 2 arccos(7/25) ]  = 

(625/ 128) * sin [ 2 arccos (7/25) ]  =

(625 / 128) * (336/625) =

336/128    =

( 21 /  8  )   units^2

2) Area of the isosceles triangle ABC =12 [By bisecting BC, we have, 3, 4, 5 right triangle]

Since O is the circumcenter of triangle ABC, then:

The circumradius OC=OB=3.125. The circumradius R = a.b.c / 4*Area =5.5.6 / 4*12=3.125.

Then the area of the isosceles triangle OBC with sides OC=OB=3.125 and BC=6 and h=0.875

=2 5/8 units^2