Proof:
{
r
θ
=
L
1
R
θ
=
L
2
R
=
r
+
h
\begin{cases} r\theta=L_1 \\ R\theta=L_2 \\ R = r+h \end{cases}
⎩⎪⎨⎪⎧rθ=L1Rθ=L2R=r+h
==>
r
=
L
1
h
L
2
−
L
1
r=\frac{L_1h}{L_2-L_1}
r=L2−L1L1h
A ( D B C E ) = A ( A B C ) − A ( A D E ) = 1 2 θ R 2 − 1 2 θ r 2 = 1 2 L 2 R R 2 − 1 2 L 1 r r 2 = 1 2 L 2 R − 1 2 L 1 r = 1 2 L 2 ( r + h ) − 1 2 L 1 r = 1 2 ( L 2 ( L 1 h L 2 − L 1 + h ) − L 1 L 1 h L 2 − L 1 ) = 1 2 L 2 L 1 h + L 2 2 h − L 2 L 1 h − L 1 2 h L 2 − L 1 = 1 2 ( L 1 + L 2 ) h A_{(DBCE)}=A_{(ABC)}-A_{(ADE)}=\frac{1}{2}\theta R^2 - \frac{1}{2}\theta r^2=\frac{1}{2}\frac{L_2}{R} R^2 - \frac{1}{2}\frac{L_1}{r} r^2=\frac{1}{2}{L_2}R - \frac{1}{2}{L_1}r=\frac{1}{2}{L_2}(r+h) - \frac{1}{2}{L_1}r=\frac{1}{2}({L_2}(\frac{L_1h}{L_2-L_1}+h) - {L_1}\frac{L_1h}{L_2-L_1})=\frac{1}{2} \frac{L_2 L_1 h+L_2^2h-L_2L_1h-L_1^2h}{L_2-L_1}=\frac{1}{2}(L_1+L_2)h A(DBCE)=A(ABC)−A(ADE)=21θR2−21θr2=21RL2R2−21rL1r2=21L2R−21L1r=21L2(r+h)−21L1r=21(L2(L2−L1L1h+h)−L1L2−L1L1h)=21L2−L1L2L1h+L22h−L2L1h−L12h=21(L1+L2)h
扫一扫
2620
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
