CPhill

Note that angle PAD = 150

AD = AP

So

angles ADP  and APD are equal

APD =  (180 -150) /2    = 15°

Therefore  angle BPD    = 60  -15    = 45°

The scale factor  of the smaller prism to the larger  =  36/48  = 3/4

Volume of smaller prism =  Volume of larger prism * (scale factor)^3 =    576 * (3/4)^4   =   243

The ratio of their lateral areas = (scale factor)^2 =  (3/4)^2  =  9 / 16

                     A

B  8     D      x         C     24             E

[ABC] = 4

So

4 = (1/2) (8 + x) (h)

8 / (8 + x)  = h    (1)

And   

[ADE ] = 9

So

8 =(1/2) (24 + x) h

16 / (24 +x)  = h   (2)

Equate (1) (2)

8 /(8 + x)  = 16 / (24 + x)

1/ (8 + x)  = 2 /(24 + x)

2 (8 + x) = 24 + x

16 + 2x  = 24 + x

x = 8

h = 8 / ( 8 + 8)    = 1/2

So

[ ACD ]  = (1/2) ( 8) (1/2)  =  2

Here's the same  problem with just  some  different numbers

https://web2.0calc.com/questions/triangles_9

See if you can work through it......if you get stuck, let me know....

angle ABP + angle BPA  = 90

angle DPQ + angleBPA = 90

So

angle ABP  =  angle DPQ

tan ABP  = 1/3

So  tan DPQ  = 1/3

So 

 1/3 = DQ/2

DQ = (2/3)

So

[ PDQ ]  =  (1/2) PD * DQ  =  (1/2) (2) (2/3)  =  2/3

If C = 90°and the triangle is isosceles, then the other two angles each  = 45°

So B can only = 45°

(x)  + (2x + 10)  + 3(2x +10)  - 10 =  180

x + 2x + 10 + 6x + 20  = 180

9x + 30  =180

9x = 150

x = 150/9  = 50 / 3

Largest angle =     6(50/3) + 20  =  120°

bh =  28

h =  28/b

So.....by the Pythagorean Theorem 

(b/2)^2  + (28/b)^2   =7^2

b^2/4 + 784 / b^2  = 49

b^4 /4 + 784  = 49b^2

b^4/4 - 49/b^2  + 784  = 0

The  only  positive value for b ≈ .25

Only possible perimeter ≈    7 + 7 + .25  ≈  14.25

Thanks for catching my error....corrected  !!!

Each side of the similar triangle is twice  as long as in the orignal triangle

So  perimeter  = ( 16 + 20 + 20)  =  56