弧形面积计算题：已知弧形ABC，其弦长L=3.24; 矢高h=0.39.求：弧形ABC的面积。解：先求弧半径R 。在直角三角形⊿OKC中，R2=（L/2）2 （R – h）2=R2－2hR h2 L2 / 4.得：R = ( h2 L2/4) / 2h =h / 2 L2 / 8h = 0.39 / 2 3.242 /( 8×0.39) = 3.559615385. ∠BOC = arcsin(L/2)/R = arcsin1.62/3.559615385 = 27.07171274°. 园心角∠AOC = 2∠BOC = 2×27.07171274°= 54.14342548°.再求扇形ABCO面积：扇形ABCO面积 = 3.1416 R2×54.14342548°/360°= 5.986862047.弧形ABC的面积 = 扇形ABCO面积 - 三角形⊿AOC面积 = 5.986862047 – L（R-h）/2 = 5.986862047 – 3.24×(3.559615385–0.39)/2 = 0.852085123.(画好的图复制不过来)
弧形ABC的面积 = 扇形ABCO面积 - 三角形⊿AOC面积
= 5.986862047 – L（R-h）/2
= 5.986862047 – 3.24×(3.559615385–0.39)/2
= 0.852085123.
弧长 = 园周长×园心角∠AOC / 360°
= 2×3.1416 R×54.14342548°/360°
= 2×3.1416 ×3.559615385×54.14342548°/360°
= 3.363769059.
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