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八年级数学解答题一般
题目
定义:若以三条线段aa,bb,cc为边能构成一个直角三角形,则称线段aa,bb,cc是勾股线段组.

(1)(1)如图①,已知点MM,NN是线段ABAB上的点,线段AMAM,MNMN,NBNB是勾股线段组.若AB=12AB=12,AM=3AM=3,求MNMN的长;
(2)(2)如图②,ABC\triangle ABC中,A=17\angle A=17^{\circ},B=28\angle B=28^{\circ},边ACAC,BCBC的垂直平分线分别交ABAB于点MM,NN,求证:线段AMAM,MNMN,NBNB是勾股线段组;
(3)(3)如图③,在等边ABC\triangle ABC,PPABC\triangle ABC内一点,线段APAP,BPBP,CPCP构成勾股线段组,CPCP为此线段组的最长线段,求APB\angle APB的度数.
知识点:三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)(1)AB=12\because AB=12AM=3AM=3
MB=9\therefore MB=9
MN=xMN=x
①若MNMN为直角边,
32+x2=(9x)2\therefore 3^{2}+x^{2}=\left(9-x\right)^{2}
x=4\therefore x=4
②若MNMN为斜边,
32+(9x)2=x2\therefore 3^{2}+\left(9-x\right)^{2}=x^{2}
x=5\therefore x=5
MN=4\therefore MN=455
(2)(2)证明:如图②,连结CMCMCNCN

\becauseACACBCBC的垂直平分线分别交ABAB于点MMNN
AM=CM\therefore AM=CMBN=CNBN=CN
A=ACM=17\therefore \angle A=\angle ACM=17^{\circ}B=NCB=28\angle B=\angle NCB=28^{\circ}
A+ACM+B+BCN=90\therefore \angle A+\angle ACM+\angle B+\angle BCN=90^{\circ}
MCN=90\therefore \angle MCN=90^{\circ}
CM\therefore CMCNCNMNMN为勾股线段组,
AM\therefore AMMNMNNBNB为勾股线段组;
(3)(3)如图③,将ABP\triangle ABP绕点AA逆时针旋转6060^{\circ}得到ACH\triangle ACH
AP=AH\because AP=AHPAH=60\angle PAH=60^{\circ}

AHP\therefore \triangle AHP是等边三角形,
AHP=60\therefore \angle AHP=60^{\circ}
AP\because APBPBPCPCP为勾股线段组,BP=CHBP=CHPH=APPH=AP
PH2+CH2=PC2\therefore PH^{2}+CH^{2}=PC^{2}
PHC=90\therefore \angle PHC=90^{\circ}
APB=AHC=150\therefore \angle APB=\angle AHC=150^{\circ}.

解析

(1)(1)AB=12\because AB=12AM=3AM=3
MB=9\therefore MB=9
MN=xMN=x
①若MNMN为直角边,
32+x2=(9x)2\therefore 3^{2}+x^{2}=\left(9-x\right)^{2}
x=4\therefore x=4
②若MNMN为斜边,
32+(9x)2=x2\therefore 3^{2}+\left(9-x\right)^{2}=x^{2}
x=5\therefore x=5
MN=4\therefore MN=455
(2)(2)证明:如图②,连结CMCMCNCN

\becauseACACBCBC的垂直平分线分别交ABAB于点MMNN
AM=CM\therefore AM=CMBN=CNBN=CN
A=ACM=17\therefore \angle A=\angle ACM=17^{\circ}B=NCB=28\angle B=\angle NCB=28^{\circ}
A+ACM+B+BCN=90\therefore \angle A+\angle ACM+\angle B+\angle BCN=90^{\circ}
MCN=90\therefore \angle MCN=90^{\circ}
CM\therefore CMCNCNMNMN为勾股线段组,
AM\therefore AMMNMNNBNB为勾股线段组;
(3)(3)如图③,将ABP\triangle ABP绕点AA逆时针旋转6060^{\circ}得到ACH\triangle ACH
AP=AH\because AP=AHPAH=60\angle PAH=60^{\circ}

AHP\therefore \triangle AHP是等边三角形,
AHP=60\therefore \angle AHP=60^{\circ}
AP\because APBPBPCPCP为勾股线段组,BP=CHBP=CHPH=APPH=AP
PH2+CH2=PC2\therefore PH^{2}+CH^{2}=PC^{2}
PHC=90\therefore \angle PHC=90^{\circ}
APB=AHC=150\therefore \angle APB=\angle AHC=150^{\circ}.

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