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Note that we  have   25  ways  to  have the combo   A_

Then 24 ways to have the combo  B_

So....the number of  combos is just the sum of the first  25  positive integers  =  

(25 * 26)  / 2  = 

25 * 13  =

325  ways

Since AB  is 3 units longer than CD.....then the height of the trapezium  = 4  =AD

We have that

34  = (1/2)(4)  ( CD  + 3  + CD)

17  =  3 + 2CD

14  =  2CD

7  = CD 

10 = AB

4 = AD

5  = BC

So.....the perimeter  =   CD + AB  +  AD  +  BC   =  7  + 10   +  4   +  5    =  26

thxd dragan

thx cphill

Very nice Omi!   

Good job, Dragan   !!!!

Let  the bottom left vertex of the larger squaree= (0,0)

Let the top right vertex of the larger square  = (12,12)

Let the bottom right vertex  of the smaller square  = (18, -6)

One side of the  triangle =  12√2  =   √288 

Another side  of  the  triangle  =  √ (18^2 + 6^2)  = √ (360)   

And the remaining side of the triangle  = √  [ ( 18 -12)^2  + (-6 - 12)^2 ]  = √360  

And the height of this isoceles triangle =  √ [ (√360)^2  - (√72)^2 ]   =  √288

So....the shaded  area  = 

(1/2) (12√2) (√288)  =

6 √2 * √288  =

6 √2 * 12√2  =

144 units^2

222^555 =1.6822898324191972919276882926373e+1302
333^444 = 9.2308071292417644488856163205835e+1119
444^333 = 3.7890387621992308448721621423179e+881
555^222 = 1.7101816934244160638009143227707e+609
Can you tell which is the largest?

Smaller square side                  a = 1

Equilateral triangle side             s = 1

Equilateral triangle height         h =  sqrt [ s² - ( s/2)² ]       h = sqrt [ 1 - (1/2)² ]  = 0.866

The area of a larger square      A = 2( h + a/2 )²                 A = 2(0.866+0.5)²  =  3.732 u²                                        

What is the radius of the small, red circle?

The height, H,  of one of the equilateral triangles   can be found as

tan 60   =   H /(1/2)

√3/2  = H

The diagonal of the square  =   2H  + 1 =   √3  +  1

The side of the larger  square = diagonal / √2  =     [ √3 + 1  ]  / √2

So....the area  of the larger square  =  

[ √3 + 1 ] ^2 / 2     =    [ 3 + 2√3 + 1' / 2   =  [4 + 2√3 ] / 2   =   [ 2 + √3 ]  units^2  ≈  3.732 units^2 

Let w = a + bi and z = c + di.  Then

\(\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\)

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\(\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\)

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (w + z)/(1 + wz) is real.

See the image

AB = 12       BC  = 15       so     AC =   9

Since BD bisects   angle ABC,  then

AD / DC =  AB/ BC

AD / DC = 12/15

AD / DC =  4/5

So   AD =  (4/9)(9) =  4

So    D  =(4,0)    and  DC =  AC - AD  =  9 - 4    =  5

And the distance from B to D  =  sqrt  (12^2  + 4^2 )  =  sqrt  (160)

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

BC^2  =  DC^2  + BD^2   -2(DC* BD)cos(BDC)

15^2  = 5^2  +  160  -  2 ( 5sqrt(160)) cos(BDC)

40 =  -10sqrt (160) cos (BDC)

-40 /  (10 sqrt (160) )  =  cos (BDC)

-4 / [ sqrt (160) ]  =cos (BDC)

-4sqrt (160)  / 160  = cos (BDC)

-sqrt (160) / 40  = cos (BDC)

- sqrt (10)  / 10  =  cos (BDC)

26 C 2 =325 ways such as:AB, AC, AD......AX, AY, AZ.......25 combinations. Then it continues with B such as BC, BD, BE.........BX, BY, BZ........for 24 combinations ........and so on.

sry thats not right

  x² + x – 12  factors to   (x + 4) (x – 3) 

–x² –4x + 21  factors to   (x + 7) (–x + 3)    call this (x + 7) (x – 3) (–1)

Substitute factored expressions in original.

Simultaneously invert divider and multiply.           (x + 4) (x – 3)             (x + 7)

                                                                            (x + 7) (x – 3) (–1)      (x + 4)

Do a bunch of cancelling terms.                           (x + 4) (x – 3)             (x + 7)

                                                                             (x + 7) (x – 3) (–1)      (x + 4)

If I did everything right, the simplified answer is     –1          

_.

Draw the triangle ABC, Plot D on AC and draw a line from angle ABC to point D, Thus, ABD=DBC

In the right angle triangle ABC, we know that AB=12, BC=15 therefore, AC=9 

We have all sides so we can use either sine or cosine or tangent to find angle ABC 

let's use cosine 

Let, angle ABC=y

cos(y)=12/15

y=cos^-1(12/15) = 36.869897645844 

Since BD bisects ABC 

Therefore BDC=BDA as we said earlier and we know that DBC+DBA=ABC thus, BDC= 1/2ABC 

so BDC=(1/2)36.869897645844 = 18.434948822922

We know 2 angles in a triangle so the third angle 

180-(36.869897645844+90)=53.130102354156 (Angle BCA)

In triangle BDC, we know 2 angles which are Angle DBC=18.434948822922 and angle BCA=53.1301023541456

Therefore angle BDC=180-(53.1301023541456+18.434948822922) = 108.4349488229324

cos(108.4349488229324) = -0.316227766017

Note finding angle BDC is easier if we just added angle BAD+DBA since BDC is exterior angle of triangle ABD. 

Combining the areas of all the regions, I'm getting that the area of the large square is 3 + 2*sqrt(3).

Confirmation from Wolfram:

https://www.wolframalpha.com/input/?i=systems+of+equations+calculator&assumption=%22FSelect%22+-%3E+%7B%7B%22SolveSystemOf4EquationsCalculator%22%7D%7D&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation1%22%7D+-%3E%222x%2By%3D14%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation2%22%7D+-%3E%22L%2BL%2Bz%3D20%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation3%22%7D+-%3E%22x%2By%2Bz%3D12%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation4%22%7D+-%3E%22x%2BL%2Bz%3D18%22

2x+y=14 (1)

L+L+z=20 (2)

x+y+z=12 (3)

x+z+L=18 (4)

\(0L+2x+y+0z=14\)

\(2L+0x+0y+z=20\)

\(0L+x+y+z=12\)

\(L+x+0y+z=18\)

Add the 4 equations

\(3L+4x+2y+3z=64\)

3(L+z)+2(2x+y)=64   I factorising by grouping.

We know from (1) that 2x+y=14

so

3(L+z)+2(14)=64

3(L+z)+28=64    I subtract 28

3(L+z)=36            I divide by 3

L+z=12 (5)

So we can find the value of L from (2)

L+L+z=20

we know that L+z=12 from (5)

L+12=20

L=8 

Well also from (2) since we knew that L=8 we can now z

8+8+z=20

z=20-16

z=4

from (4) we know that x+z+L=18

so

x+4+8=18

x=18-12

x=6

The question may end here as you asked "Find x+z" we know x and we know z so x+z=6+4=10 , the answer is 10.

However, let's contuine to get the value of y.

from (1) we know that x+x+y=14

we know that x=6

thus,

6+6+y=14

y=14-12

y=2

So,

L=8

x=6

y=2

z=4

1 hectare is 100 meters by 100 meters = 10000 metries square 

so

12 hectare = 120 000 square metres 

The perimeter of the trapezium is  26 units.    

No, you're not missing anything, it's E.

If it is 2D....let's label the figure like this:

△ABE  is an equilateral triangle, so every interior angle is 60°, and so

m∠EAB  =  60°

m∠ABE  =  60°

We are given that

m∠EBC  =  90°

Now we can determine that

m∠ABC  =  60° + 90°

m∠ABC  =  150°

△ABC is an isosceles triangle. so base angles are congruent, and so

m∠BCA  =  m∠BAC

The sum of the measures of the interior angles in a triangle is 180°, so

m∠BAC  +  m∠BCA  +  m∠ABC  =  180°

                                                                     Substitute  m∠BAC  in for  m∠BCA  and  150°  in for  m∠ABC

m∠BAC  +  m∠BAC  +     150°    =   180°

                                                                     Combine like terms

2(m∠BAC)  +  150°  =  180°

                                               Subtract  150°  from both sides of the equation.

2(m∠BAC)  =  30°

                                               Divide both sides of the equation by  2

m∠BAC  =  15°

In the same way, we can determine that

m∠EAD  =  15°

Then

m∠DAC   =   m∠EAB  - m∠EAD - m∠BAC   =   60° - 15° - 15°   =   30°