如圖,△ABC中,∠C=90°,AB=10cm,BC=6cm,動點P從點C出發,以每秒2cm的速度按C→A的路...
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問題詳情:
如圖,△ABC中,∠C=90°,AB=10 cm,BC=6 cm,動點P從點C出發,以每秒2 cm的速度按C→A的路徑運動,設運動時間為t秒.
(1)出發2秒時,△ABP的面積為 cm2;
(2)當t為何值時,BP恰好平分∠ABC?
【回答】
(1)12.······························· 3分
(2)解:過點P作PG⊥AB於G,則∠BGP=90°.
∵∠C=90°,
∴∠BGP=∠C.···························· 4分
∵BP平分∠ABC,
∴∠CBP=∠ABP.··························· 5分
又∵BP=BP,
∴△BCP≌△BGP.··························· 6分
∴BG=BC=6,PG=PC=2t.
∴PA=8-2t,AG=10-6=4.······················ 8分
在Rt△APG中, AG2+PG2=AP2.
∴42+(2t)2=(8-2t)2 ························· 9分
解得t=.····························· 10分
(説明:用面積法求解類似給分)
知識點:勾股定理
題型:解答題