1 / 5

畢氏定理之應用

畢氏定理之應用. AB = DE = = 3. 2.4. CE = 0.6 + 1.8 =2.4. CD = =1.8. 0.6. 1.8. 畢氏定理之應用 1. 有一梯子橫靠在牆上，原先梯子的頂端 (A 點 ) 離牆角 (C 點 ) 2.4 公尺，底端 (B 點 ) 離牆腳 1.8 公尺，若梯子的底端不慎往後滑了 0.6 公尺，則梯子的頂端會往下滑動多少公尺？. A. 觀看動畫. 標示長度. 計算. Ｄ. 梯頂下滑： 2.4 － 1.8 = 0.6. E. B. C. 300.

hunter-emerson
Download Presentation

畢氏定理之應用

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 畢氏定理之應用

  2. AB = DE = = 3 2.4 CE = 0.6 + 1.8 =2.4 CD = =1.8 0.6 1.8 畢氏定理之應用1 有一梯子橫靠在牆上，原先梯子的頂端 (A點) 離牆角 (C點) 2.4 公尺，底端 (B點) 離牆腳 1.8 公尺，若梯子的底端不慎往後滑了 0.6 公尺，則梯子的頂端會往下滑動多少公尺？ A 觀看動畫 標示長度 計算 Ｄ 梯頂下滑： 2.4 － 1.8 = 0.6 E B C

  3. 300 400 距離為：　 = 500 畢氏定理之應用2 若小明想要由家裡出發，沿著街道到達學校時，他必須先往北走 400 公尺，再向東走 300 公尺，則小明家與學校的直線距離為何？ 學校 觀看動畫 標示長度 計算 北 小明家

  4. B A C AB = = 5 畢氏定理之應用3 試求座標平面上A(1,1)、B(4,5)兩點間的距離為何？ 觀看動畫 計算

  5. PR = RQ = R(c,b) PQ = 畢氏定理之應用－直角座標平面上兩點之距離公式 試求座標平面上P (a,b)、Q (c,d) 兩點間的距離為何？ d Q(c,d) 觀看動畫 計算 c a b P(a,b)

More Related