Dragan

Thanks, Melody! That's how we did 50+ years ago.  

AB and CD are two parallel chords of a circle such that AB = 24 cm and CD= 10 cm. If the radius of the circle is 13 cm , find the distance between the chords.

The distance between 2 chords is 17 cm

NO = 5/(1+sqrt(2)) = 2.071068294

MO = r / sin(45) = 2.92893287

MN = MO - NO =0.857864576

Proportion:           MN  :  NO  =  MT  :  TR

Radius of circle "M" is:      TR = 12.071068 m   

CH is an altitude of the triangle ABC, m∠A = 60°, m∠B = 45°. Circle with the diameter CH intersects sides AC and BC at points N and M respectively. Find the area (exact value) of the quadrilateral CMHN if AN = 1.

Area of the quadrilateral CMHN = 5.598 u²  

My mistake... it's 65°  

You can do it; it's easy.

Angle JGH = 180 - 2 * 62 = 56 degrees

In the diagram below, ABCD is a parallelogram with AB=42 cm. Arc AD is a semi-circle. The area of the parallelogram is 504cm2, and angle ADC = 60 degrees. Find the area of the shaded region in cm2.

AB = 42cm       

Area = 504 cm²

Angle ADC = 60°

504 / 42 = 12

Semi-circle radius  r =  [12 / sin(60°)] /2    r = 6.928

Shaded region area    A_r = ( r² pi ) /2 = 75.398 cm²    

The area of the inscribed equilateral triangle is 75/4*sqrt(3)​​. Find the perimeter of circumscribing square to the nearest whole number.

The ratio of the area of an equilateral triangle to its circumcircle's area is:   1.299 : pi  or  0.413496

Area of the triangle is        75/4*sqrt(3) = 32.47595 u²

Area of the circle is            32.47595 / 0.413496 = 78.54 u²

Raduis of the circle is        r = sqrt( 78.54 / pi ) = 5

Perimeter of a square is     P = 8r = 40 units