1.
Here's a pic :
Call the intersection of BC and B'A' , N
Note that triangle BMN is similar to triangle BCA
So
MN/ BM = AC / BC
MN / (5/2) = 3/4
MN = 15/8
So....the area of triangle BMN = (1/2) (product of the legs) = (1/2)(BM)(MN) = (1/2)(5/2)(15/8) = 75/32
So BN = sqrt [ BM^2 + MN^2] = sqrt [ (5/2)^2 +(15/8)^2 ] = 25/8
So B'N = B'M - MN = (5/2) - (15/8) = 20/8 - 15/8 = 5/8
So A'N = A'M + MN = 5/2 + 15/8 = 20/8 + 15/8 = 35/8
And call the intersection of BC and B'C', P
Because BC is parallel to A'C'.....then
A'N / A'B' = C'P / C'B'
(35/8) / 5 = C'P /4
35/40 = C'P /4
C'P = 3.5
Then B'P = B'C' - C'P = 4 -3.5 = .5
So....using the Pythagorean Theorem....NP = sqrt [B'N^2 - B'P^2] = sqrt [ (5/8)^2 - (4/8)^2 ] = 3/8
Then BP = BN + NP = 25/8 + 3/8 = 28/8 = 7/2
And call the intersection of BA and B'C', Q
And because PQ is parallel to AC
Then BP / BC = PQ/CA
(7/2) / 4 = PQ/ 3
(7/8) = PQ/3
(21/8) = PQ
So the area of triangle BPQ = (1/2) (BP)(PQ) = (1/2) (7/2) (21/8) = 147/32
So... the shaded area = [area of triangle BPQ ] - [area of triangle BMN ] =
[ 147 - 75 ] / 32 = 72/32 = 9/4 = 2.25 units^2