Can someone help me with this?

Step-by-step explanation:

Quadrilateral EFGH is a rectangle.

Therefore, measure of all interior angle of the rectangle = 90°

7). m∠FEG + m∠GEH = 90°

    57° + m∠GEH = 90°

    m∠GEH = 90° - 57°

    m∠GEH = 33°

Diagonal of the rectangle are equal in measure and bisect each other.

8). m(FH) = m(FK) + m(KH)

               = 32 + 32 [Since, FK ≅ KH]      

               = 64

9). m∠FGH + m∠EFG = 90° + 90°

                                   = 180°

MNOP is a Rhombus, so the diagonals will be perpendicular and bisect each other.

10). m∠MRN = 90°

11). m(PR) = 12

    Therefore, m(RN) = 12

"Diagonals of a rhombus bisect the opposite angles"

12). m(∠PON) = 124°

   Therefore, m(∠POM) = 124°

"ABCD is an isosceles Trapezoid"

13). Therefore, ∠C ≅ ∠D

14). AB║CD

15). AD ≅ BC

answer: about 22.67

step-by-step explanation:

other answer is correct. i had used the radius of the radius somehow. a website: