civonamzuk

It's very simple.                                                                                                                                                       

Two identical triangles with side lengths of 125 cm, 132 cm, and 37 cm form a parallelogram ABCD.

Heron's formula gives the area of a triangle when the length of all three sides is known. 

A = sqrt[s(s - a)(s - b)(s - c)]               ( s => semiperimeter )    s = 1/2(a + b + c)

Hello, jugoslav!

F and F' would "share the same point" in any triangle whose angles A and B are congruent (equilateral, isosceles triangles).

correction... 

Angle BCA = 180 - 123 = 57º

Circle area     A = 63² (22/7) = 12474 cm²

Sector angle     12474 / 360 = 4158 / x          x = 120          

Angle = 120º

20(20 + x) = 31²

400 + 20x = 961

20x = 561

x = 28.05

N = 86

86/9 = 9  R5

86/7 = 12  R2

86/5 = 17  R1

In the figure below, ABCD is a parallelogram, AD = 18 and DC = 27. Segments EG and FH are perpendicular to sides of the parallelogram as shown, and EG = 16. Determine FH.

AD ≠ EG 

FH = (27 * 16) / 18

FH = 24

A = ab + c(d + b + e) ft²

(9x + 1) + (5x + 12) + (10x - 37) = 360

24x = 384

x = 16

Angle BCA = 180 - (10x - 37)

Angle BCA = 123º

perimeter = 1879.6 mm