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these are just two semi-circles, one with a diameter of 22 and the other with a diameter of 30 - 22, or 8, and therefore their radius are 11 and 4 respectively.
Answer:
A = 20sinθ(6 + 5 cosθ) cm²
Step-by-step explanation:
Drop perpendiculars DE and CF to AB.
Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.
The formula for the area of the trapezium is
A = ½(a + b)h
DE = 10sinθ
AE = 10cosθ
BF = 10cosθ
EF = CD = 12 cm
AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ
A = ½(a + b)h
= ½(12 +12 + 20 cosθ) × 10 sinθ
=(24 + 20 cosθ) × 5 sinθ
= 4(6 + 5cosθ) × 5sinθ
= 20sinθ(6 + 5 cosθ) cm²
The gcf he can make is 131,313 packages
