The sector OAB is one-quarter of a circle with a radius of 8cm. C is the mid-point of OA in which a semicircle ODA is drawn with C as its center.A line CDE, which is perpendicular to OA, is drawn at C. Find the shaded area.
explain how to do plz.........
& thx!!!!!!!!!!!!!!!!!!!!!
http://img100.imageshack.us/my.php?image=aabbtj1.png
中四a maths 一問 急!!!
2007-12-14 5:50 am
回答 (2)
2007-12-14 6:39 am
✔ 最佳答案
Joining OEcos∠COE = OC/OE = 4/8
∠COE = 60°
Area of sector OBE
= π(8)² x (90° - 60°)/360°
= 16π/3 cm²
Area of ΔOCE
= (4 x 8sin60°)/2
= 8√3 cm²
Area of sector OCD
= π(4)² x 90°/360°
= 4π cm²
The shaded area
= 16π/3 + 8√3 - 4π
= (4π/3 + 8√3) cm²
2007-12-15 9:45 pm
connect O to E,
cos angle EOC=CO/EO=4/8=60 degree
CE=OEx(sin60)=8xsin60=4(square root 3)
the shaded area of ODE
=1/2x4x4x(square root)3-1/4x(pie)x4^2
=8(square root)3-4(pie)
the shaded area of ODEB
=30/360x8^2x(pie)+8(square root)3-4(pie)
=16/3(pie)+8(square root)3-4(pie)
=4/3(pie)+8(square root)3
cos angle EOC=CO/EO=4/8=60 degree
CE=OEx(sin60)=8xsin60=4(square root 3)
the shaded area of ODE
=1/2x4x4x(square root)3-1/4x(pie)x4^2
=8(square root)3-4(pie)
the shaded area of ODEB
=30/360x8^2x(pie)+8(square root)3-4(pie)
=16/3(pie)+8(square root)3-4(pie)
=4/3(pie)+8(square root)3
收錄日期: 2021-04-13 14:42:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071213000051KK03852