jugoslav

AD = BC = ED = FD = 37

EF = 2 * 37 = 74 

∠ABD ≅ ∠DBC

3x = 6x - 30

- 3x = - 30  

-x = -10  | *-1

x = 10

Angles AOD and COB are congruent

2x + 30 = x + 90

x = 60 degrees

The distance between the points (x, 21) and (4, 7) is 10√2. Find the sum of all possible values of x.

10√2  is a hypotenuse of a right triangle, the longer leg is 14.                                                                                          The short leg     a = sqrt[(10√2)² - 14²] = 2

1/      x = 4 - a = 2

2/      x = 4 + a = 6

So, the sum of all possible values of x is 8 

180 - 4x + x + 180 - 147 = 180

- 3x = 180 - 180 - 33

- 3x = - 33  | * -1

3x = 33

x = 11

Just kiddin'

Is this forum for the third graders too???

A / XYZ = 4/9  

Let an angle BAD be  β       β = 7º

1/       AD = cosβ * 3.7

2/      BD = sinβ * 3.7

3/      A_(ΔABD) = (AD * BD) / 2

4/      A_(ΔADC) = A(ΔABD) + 22.4

5/     CD = (2 * A_(ΔADC)) / AD

6/     b = sqrt( AD² + CD² )   

 A(0, 3)      B(12, 0)       origin O       center of a circle  C   

1/     Line segment  AB = sqrt(AO² + BO²)

 2/    Angle (ABO) = arctan(AO / BO)              ∠ABO = ∠BAC

3/     Radius  AC = (AB/2) / cos∠BAC