Question

PLEASE HELP ME ASAP PIC ATTACHED

Answer 1

Answer: 90,75. cos 55°. sin55° ≈ 42,64

through E, draw EK perpendicular to BC (F ∈ BC)

because E is the midpoint of AB => BE = 11/2 = 5.5

ΔBEF is a right triangle at K

=> sin 55° = EK/BE

=> EK = 5,5.sin 55°

ΔABC has E and F is the midpoints of AB and AC

=> EF is  the median line of the triangle

=> EF = 1/2.BC

through A, draw AD perpendicular to BC

=> ΔABD is a right triangle at D

=> BC = 11.cos 55°.2 = 22.cos 55°

=> EF = 11.cos 55°

=>the area of the trapezium BEFC is:

$\frac{5,5.sin55(11.cos55+22.cos55)}{2}=90,75.cos55.sin55$

Step-by-step explanation: