2 Problems

1.  

The side of the smaller square  = 1 cm   and the side of the larger square  = 4 cm

Let the distance from  the top right of the smaller square to the intersection of AB with the side of the larger square  = x

So  the small triangle formed above the smaller square has a hypotenuse of   √[ 1 + x^2]

And  the larger triangle formed at the "top" part  of the larger square has  legs of  4  and [ (4) - (1 + x)]  =  (3 -x)    so    the hypotenuse  of this triangle  is  √  [ 4^2  + (3 - x)^2 ]  =

√ [  16  + 9 - 6x  + x^2 ]  =  √ [ x^2 - 6x + 25 ]

And the triangles are similar  such that

1 / √[ 1 + x^2]  =  4 /  √ [ x^2 - 6x + 25 ]      cross-multiply

√ [ x^2 - 6x + 25 ]  =  4 √[ 1 + x^2]          square both sides

x^2  - 6x + 25   =  16 (x^2 + 1)

x^2  - 6x + 25  =  16x^2  + 16

15x^2 + 6x - 9  =  0

5x^2  + 2x - 3  =  0

(5x - 3) (x + 1)  =  0

Setting each factor to 0 and solving for x produces   x  = 3/5  = .6  and x  = -1

But  x  is a length so it can't be negative

So  x  =  3/5 cm = .6cm

So....the length of AB  is

√[ 1 + (.6)^2]   +   √ [ (.6)^2 - 6(.6) + 25 ]  =

√ [ 1.36]  +  √ [ 21.76 ]  ≈ 5.8  cm    

2. We can  find AB  thusly

sin (60)  = AB / 24

24 * √3 /2  =  AB

12√3  = AB

And  BE  =  (1/2)  of   AE  = 12

But ABE  and  BCE form similar triangles

So....if  BE  is  (1/2) of AE

Then BC  is  (1/2)  of AB  = 6√3

And CE  is  1/2 of BE  =  6

And triangle DCE is similar to triangle CBE

So since CE  is (1/2) of BE....then...

ED  is(1/2)  of CE  = 3     and

DC  is (1/2) BC  = 3√3

So  the perimeter of   ABCD  =

AB + BC + CD  + DE + EA  =

12√3  + 6√3  + 3√3  + 3  + 24  =

 [ 21√3  +   27 ] cm  ≈  63.4 cm

If I haven't made any errors  !!!!

First one: slightly different approach...see diagram   the 1's and 4's were given in the question for the squares,