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八年级数学填空题一般
题目
如图,已知ABC\triangle ABC的面积是6060,请完成下列问题:

(1)(1)如图11,ABC\triangle ABC中,若ADADBCBC边上的中线,则ABD\triangle ABD的面积______ACD\triangle ACD的面积(填">\gt"、"<\lt"或"==");
(2)(2)如图22,若CDCDBEBE分别是ABC\triangle ABCABABACAC边上的中线,求四边形ADOEADOE的面积可以用如下方法:连接AOAO,设SADO=xS_{\triangle ADO}=x,SAEO=yS_{\triangle AEO}=y,联想第一小问结论,通过列方程组来求四边形ADOEADOE的面积;
(3)(3)如图33,AD:DB=1:3AD:DB=1:3,CE:AE=2:3CE:AE=2:3,请求出四边形ADOEADOE的面积.
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)AD\left(1\right)\because ADBCBC边上的中线,
ABD\therefore \triangle ABDACD\triangle ACD是等底等高的三角形,
SABD=SACD\therefore S_{\triangle ABD}=S_{\triangle ACD}
故答案为:==
(2)CD(2)\because CDABAB边上的中线,
ADO\therefore \triangle ADOBDO\triangle BDO是等底等高的三角形,
SADO=SBDO\therefore S_{\triangle ADO}=S_{\triangle BDO}
BE\because BEABC\triangle ABCACAC边上的中线,
CDO\therefore \triangle CDOAEO\triangle AEO是等底等高的三角形,
SCEO=SAEO\therefore S_{\triangle CEO}=S_{\triangle AEO}.
SADO=xS_{\triangle ADO}=xSAEO=yS_{\triangle AEO}=y,则SBDO=xS_{\triangle BDO}=xSCEO=yS_{\triangle CEO}=y,由题意得:
SABE=12SABC=30S_{\triangle ABE}=\frac{1}{2}S_{\triangle ABC}=30SADC=12SABC=30S_{\triangle ADC}=\frac{1}{2}S_{\triangle ABC}=30
{2x+y=30x+2y=30\therefore \left\{\begin{array}{c}2x+y=30\\ x+2y=30\end{array}\right.
解得:{x=10y=10\left\{\begin{array}{l}x=10\\ y=10\end{array}\right.
S四边形ADOE=x+y=10+10=20\therefore S_{四边形ADOE}=x+y=10+10=20
(3)(3)如图33,连接AOAO

AD:DB=1:3\because AD:DB=1:3
SADO=13SBDO\therefore S_{\triangle ADO}=\frac{1}{3}S_{\triangle BDO}
CE:AE=2:3\because CE:AE=2:3
SCEO=23SAEO\therefore S_{\triangle CEO}=\frac{2}{3}S_{\triangle AEO}
SADO=xS_{\triangle ADO}=xSCEO=yS_{\triangle CEO}=y,则SBDO=3xS_{\triangle BDO}=3xSAEO=32yS_{\triangle AEO}=\frac{3}{2}y
由题意得:SABE=35SABC=36S_{\triangle ABE}=\frac{3}{5}S_{\triangle ABC}=36SADC=14SABC=15S_{\triangle ADC}=\frac{1}{4}S_{\triangle ABC}=15
{4x+32y=36x+52y=15\therefore \left\{\begin{array}{c}4x+\frac{3}{2}y=36\\ x+\frac{5}{2}y=15\end{array}\right.
解得:{x=13517y=4817\left\{\begin{array}{c}x=\frac{135}{17}\\ y=\frac{48}{17}\end{array}\right.
S四边形ADOE=SADO+SAEO=x+32y=20717\therefore S_{四边形ADOE}=S_{\triangle ADO}+S_{\triangle AEO}=x+\frac{3}{2}y=\frac{207}{17}
故四边形ADOEADOE的面积为20717\frac{207}{17}.

解析

(1)AD\left(1\right)\because ADBCBC边上的中线,
ABD\therefore \triangle ABDACD\triangle ACD是等底等高的三角形,
SABD=SACD\therefore S_{\triangle ABD}=S_{\triangle ACD}
故答案为:==
(2)CD(2)\because CDABAB边上的中线,
ADO\therefore \triangle ADOBDO\triangle BDO是等底等高的三角形,
SADO=SBDO\therefore S_{\triangle ADO}=S_{\triangle BDO}
BE\because BEABC\triangle ABCACAC边上的中线,
CDO\therefore \triangle CDOAEO\triangle AEO是等底等高的三角形,
SCEO=SAEO\therefore S_{\triangle CEO}=S_{\triangle AEO}.
SADO=xS_{\triangle ADO}=xSAEO=yS_{\triangle AEO}=y,则SBDO=xS_{\triangle BDO}=xSCEO=yS_{\triangle CEO}=y,由题意得:
SABE=12SABC=30S_{\triangle ABE}=\frac{1}{2}S_{\triangle ABC}=30SADC=12SABC=30S_{\triangle ADC}=\frac{1}{2}S_{\triangle ABC}=30
{2x+y=30x+2y=30\therefore \left\{\begin{array}{c}2x+y=30\\ x+2y=30\end{array}\right.
解得:{x=10y=10\left\{\begin{array}{l}x=10\\ y=10\end{array}\right.
S四边形ADOE=x+y=10+10=20\therefore S_{四边形ADOE}=x+y=10+10=20
(3)(3)如图33,连接AOAO

AD:DB=1:3\because AD:DB=1:3
SADO=13SBDO\therefore S_{\triangle ADO}=\frac{1}{3}S_{\triangle BDO}
CE:AE=2:3\because CE:AE=2:3
SCEO=23SAEO\therefore S_{\triangle CEO}=\frac{2}{3}S_{\triangle AEO}
SADO=xS_{\triangle ADO}=xSCEO=yS_{\triangle CEO}=y,则SBDO=3xS_{\triangle BDO}=3xSAEO=32yS_{\triangle AEO}=\frac{3}{2}y
由题意得:SABE=35SABC=36S_{\triangle ABE}=\frac{3}{5}S_{\triangle ABC}=36SADC=14SABC=15S_{\triangle ADC}=\frac{1}{4}S_{\triangle ABC}=15
{4x+32y=36x+52y=15\therefore \left\{\begin{array}{c}4x+\frac{3}{2}y=36\\ x+\frac{5}{2}y=15\end{array}\right.
解得:{x=13517y=4817\left\{\begin{array}{c}x=\frac{135}{17}\\ y=\frac{48}{17}\end{array}\right.
S四边形ADOE=SADO+SAEO=x+32y=20717\therefore S_{四边形ADOE}=S_{\triangle ADO}+S_{\triangle AEO}=x+\frac{3}{2}y=\frac{207}{17}
故四边形ADOEADOE的面积为20717\frac{207}{17}.

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