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13. 如图,在$△ ABC$中,已知$AB=2$,$AD ⊥ BC$,垂足为$D$,$BD=2CD$.若$E$是$AD$的中点,求$EC$的长度.
答案
13. 解:设 $AE = ED = x,CD = y$,
则 $BD = 2y,AD = 2ED = 2x$.
$\because AD ⊥ BC$,
$\therefore ∠ ADB = ∠ ADC = 90°$.
在 $\mathrm{Rt}△ ABD$ 中,
$\because AB^2 = 4x^2 + 4y^2 = 4$,
$\therefore x^2 + y^2 = 1$.
$\therefore$ 在 $\mathrm{Rt}△ CDE$ 中,$EC^2 = x^2 + y^2 = 1$,
$\therefore EC = 1$.
则 $BD = 2y,AD = 2ED = 2x$.
$\because AD ⊥ BC$,
$\therefore ∠ ADB = ∠ ADC = 90°$.
在 $\mathrm{Rt}△ ABD$ 中,
$\because AB^2 = 4x^2 + 4y^2 = 4$,
$\therefore x^2 + y^2 = 1$.
$\therefore$ 在 $\mathrm{Rt}△ CDE$ 中,$EC^2 = x^2 + y^2 = 1$,
$\therefore EC = 1$.
14. 如图,在$△ ABC$中,$AB:BC:CA = 3:4:5$,且$△ ABC$的周长为$36\ \mathrm{cm}$. 点$P$从点$A$开始沿$AB$边向点$B$以每秒$1\ \mathrm{cm}$的速度移动,点$Q$从点$B$开始沿$BC$边向点$C$以每秒$2\ \mathrm{cm}$的速度移动. 如果它们同时出发,当点$P,Q$移动$3\ \mathrm{s}$时,$△ BPQ$的面积为多少?
答案
14. $18\ \mathrm{cm}^2$
