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题目
阅读下列材料,并解答相应问题:已知ABC\triangle ABC的面积为6060,ABABACAC边上的中线CDCDBEBE相交于点OO,如图11所示.
(1)(1)求四边形ADOEADOE的面积.
小强用了如下的方法:连接AOAO,设SBDO=xS_{\triangle BDO}=x,SCEO=yS_{\triangle CEO}=y,则SADO=xS_{\triangle ADO}=x,SAEO=yS_{\triangle AEO}=y,由题意得SABE=12SABC=30S_{\triangle ABE}=\frac{1}{2}{S}_{△ABC}=30,SADC=12SABC=30S_{\triangle ADC}=\frac{1}{2}{S}_{△ABC}=30,可列方程组{2x+y=30x+2y=30\left\{\begin{array}{l}{2x+y=30}\\{x+2y=30}\end{array}\right.,通过解这个方程组,可得四边形ADOEADOE的面积为______;
(2)(2)如图22,已知AD:BD=2:1AD:BD=2:1,CE:AE=3:1CE:AE=3:1,则四边形ADOEADOE的面积为______.
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)由{2x+y=30x+2y=30\left\{\begin{array}{l}{2x+y=30}\\{x+2y=30}\end{array}\right.,可得{x=10y=10\left\{\begin{array}{l}{x=10}\\{y=10}\end{array}\right.
S四边形ADOE=SADO+SAEO=x+y=10+10=20\therefore S_{四边形ADOE}=S_{\triangle ADO}+S_{\triangle AEO}=x+y=10+10=20.
故答案为:2020
(2)(2)如图22中,连接AOAO.

AD:BD=2:1\because AD:BD=2:1
SADO=2SBDO\therefore S_{\triangle ADO}=2S_{\triangle BDO}
CE:AE=3:1\because CE:AE=3:1
SCEO=3SAEO\therefore S_{\triangle CEO}=3S_{\triangle AEO}
SADO=xS_{\triangle ADO}=xSAEO=yS_{\triangle AEO}=y,则SBDO=12xS_{\triangle BDO}=\frac{1}{2}xSCEO=3yS_{\triangle CEO}=3y
由题意得:SABE=14SABC=15S_{\triangle ABE}=\frac{1}{4}S_{\triangle ABC}=15SADC=23SABC=40S_{\triangle ADC}=\frac{2}{3}S_{\triangle ABC}=40
可列方程组为:{x+12x+y=15x+y+3y=40\left\{\begin{array}{l}{x+\frac{1}{2}x+y=15}\\{x+y+3y=40}\end{array}\right.
解得:{x=4y=9\left\{\begin{array}{l}{x=4}\\{y=9}\end{array}\right.
S四边形ADOE=SADO+SAEO=x+y=4+9=13\therefore S_{四边形ADOE}=S_{\triangle ADO}+S_{\triangle AEO}=x+y=4+9=13.
故答案为:1313.

解析

(1)由{2x+y=30x+2y=30\left\{\begin{array}{l}{2x+y=30}\\{x+2y=30}\end{array}\right.,可得{x=10y=10\left\{\begin{array}{l}{x=10}\\{y=10}\end{array}\right.
S四边形ADOE=SADO+SAEO=x+y=10+10=20\therefore S_{四边形ADOE}=S_{\triangle ADO}+S_{\triangle AEO}=x+y=10+10=20.
故答案为:2020
(2)(2)如图22中,连接AOAO.

AD:BD=2:1\because AD:BD=2:1
SADO=2SBDO\therefore S_{\triangle ADO}=2S_{\triangle BDO}
CE:AE=3:1\because CE:AE=3:1
SCEO=3SAEO\therefore S_{\triangle CEO}=3S_{\triangle AEO}
SADO=xS_{\triangle ADO}=xSAEO=yS_{\triangle AEO}=y,则SBDO=12xS_{\triangle BDO}=\frac{1}{2}xSCEO=3yS_{\triangle CEO}=3y
由题意得:SABE=14SABC=15S_{\triangle ABE}=\frac{1}{4}S_{\triangle ABC}=15SADC=23SABC=40S_{\triangle ADC}=\frac{2}{3}S_{\triangle ABC}=40
可列方程组为:{x+12x+y=15x+y+3y=40\left\{\begin{array}{l}{x+\frac{1}{2}x+y=15}\\{x+y+3y=40}\end{array}\right.
解得:{x=4y=9\left\{\begin{array}{l}{x=4}\\{y=9}\end{array}\right.
S四边形ADOE=SADO+SAEO=x+y=4+9=13\therefore S_{四边形ADOE}=S_{\triangle ADO}+S_{\triangle AEO}=x+y=4+9=13.
故答案为:1313.

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