Question

上海初二数学练习册答案

P78的19.9(4)2.已知:AD是锐角三角形ABC的高,AC=20,BC=24,AD=16求证:△ABC是等腰三角形3.如图,已知△ABC中,LC=90°.........4.小明家准备在阳台上修建一个雨篷............就这些题目 要有过程

Answer 1

2、证明；因为AD是三角形ABC的高（已知)AD⊥BC所以 ∠ADC=∠ADB=90°在△ADC中，∠ADC=90°AD�0�5＋CD�0�5=AC�0�5(勾股定理)16�0�5+CD�0�5=20�0�5CD=12∵BC=24∴CD=DB=�0�5BC(线段中点的意义)∴DB=12(等量代换）在△ADB中，∠ADB=90°∴AD�0�5+DB�0�5=AB�0�5(勾股定理）16�0�5+12�0�5=AB�0�5AB=20∵AC=20（已知)∴AB=AC（等量代换）∴△ABC是等腰三角形 3、AC�0�5+BD�0�5-(AB�0�5+CD�0�5)=(AB�0�5-AC�0�5)-(BC�0�5-CD�0�5)在△ABC中，∠C=90°∴AB�0�5=AC�0�5+BC�0�5(勾股定理）在△BDC，，∠C=90°∴CD�0�5+BC�0�5=BD�0�5(勾股定理）即AC�0�5+BD�0�5-(AB�0�5+CD�0�5)=(AC�0�5+BC�0�5-AC�0�5)-(CD�0�5+BC�0�5-CD�0�5)∴BC�0�5-BC�0�5=0∴AB�0�5+CD�0�5=AC�0�5+BD�0�5 4、在△ABC中，∠B=90°∴b�0�5+a�0�5=AC�0�5（勾股定理）0.5�0�5+2�0�5=AC�0�5AC=2分之根号17∵S雨棚=AC×d =2分之根号17×8 =4又根号17

Answer 2

2、在直角三角形ADC中用勾股定理，解得CD=12，那么D是BC的中点，则△ADC≌△ADB,得到AB=AC；命题得证……