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八年级数学填空题一般
题目
【问题背景】利用方程解决实际问题是重要的思想方法.以面积寻找等量关系,求图形中线段的长度是解决一些几何问题的常见手段.
例如:如图11,在RtABCRt\triangle ABC中,ACB=90\angle ACB=90^{\circ},CDABCD\bot AB于点DD,若BC=3BC=3,AC=4AC=4,AB=5AB=5,求斜边ABAB上的高CDCD的长.
利用RtABCRt\triangle ABC的面积列出方程12×5×CD=12×3×4\frac{1}{2}×5×CD=\frac{1}{2}×3×4,求得CD=125CD=\frac{12}{5}.

【延伸应用】如图22,在RtABCRt\triangle ABC中,ACB=90\angle ACB=90^{\circ},BC=3BC=3,AC=4AC=4,AB=5AB=5,矩形CEDFCEDFABC\triangle ABC内,其中点DD在斜边ABAB上,EE,FF两点分别在直角边ACAC,BCBC上.
(1)(1)若矩形CEDFCEDF是正方形,则矩形CEDFCEDF的面积为______;(直接写出你的答案);(直接写出你的答案)
(2)(2)若矩形CEDFCEDF的两边之比为1:21:2,求矩形CEDFCEDF的面积.
【拓展探究】如图33,已知正方形ABCDABCD面积为1818,点EECBCB的延长线上,连接DEDE交边ABAB于点FF,若BE=22BE=2\sqrt{2},求AFAF的长.
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

延伸应用:
(1)(1)连接CDCD

\because矩形CEDFCEDF是正方形,
DF=DE\therefore DF=DE
ACB=90\because \angle ACB=90^{\circ}BC=3BC=3AC=4AC=4
SABC=12ACBC=12×4×3=6\therefore {S}_{△ABC}=\frac{1}{2}AC•BC=\frac{1}{2}×4×3=6SABC=SBCD+SACD=12BCDF+12ACDE{S}_{△ABC}={S}_{△BCD}+{S}_{△ACD}=\frac{1}{2}BC•DF+\frac{1}{2}AC•DE
12×3DF+12×4DE=6\therefore \frac{1}{2}×3•DF+\frac{1}{2}×4•DE=6,解得:DE=DF=127DE=DF=\frac{12}{7}
\therefore矩形CEDFCEDF的面积=DE2=14449=D{E}^{2}=\frac{144}{49}
故答案为:14449\frac{144}{49}
(2)(2)连接CDCD

\because矩形CEDFCEDF中两边之比为1:21:2
DF=2DEDF=2DE时,设DE=xDE=xDF=2xDF=2x
ACB=90\because \angle ACB=90^{\circ}BC=3BC=3AC=4AC=4
SABC=12ACBC=12×4×3=6\therefore {S}_{△ABC}=\frac{1}{2}AC•BC=\frac{1}{2}×4×3=6SABC=SBCD+SACD=12BCDF+12ACDE{S}_{△ABC}={S}_{△BCD}+{S}_{△ACD}=\frac{1}{2}BC•DF+\frac{1}{2}AC•DE
12×3×2x+12×4x=6\therefore \frac{1}{2}×3×2x+\frac{1}{2}×4x=6,解得:x=65x=\frac{6}{5}
\therefore矩形CEDFCEDF的面积=DEDF=x2x=2x2=2×3625=7225=DE•DF=x•2x=2{x}^{2}=2×\frac{36}{25}=\frac{72}{25}
DE=2DFDE=2DF时,设DE=2aDE=2aDF=aDF=a
ACB=90\because \angle ACB=90^{\circ}BC=3BC=3AC=4AC=4
SABC=12ACBC=12×4×3=6\therefore {S}_{△ABC}=\frac{1}{2}AC•BC=\frac{1}{2}×4×3=6SABC=SBCD+SACD=12BCDF+12ACDE{S}_{△ABC}={S}_{△BCD}+{S}_{△ACD}=\frac{1}{2}BC•DF+\frac{1}{2}AC•DE
12×3a+12×4×2a=6\therefore \frac{1}{2}×3a+\frac{1}{2}×4×2a=6,解得:x=1211x=\frac{12}{11}
\therefore矩形CEDFCEDF的面积=DEDF=2aa=2a2=2×144121=288121=DE•DF=2a•a=2{a}^{2}=2×\frac{144}{121}=\frac{288}{121}
综上,矩形CEDFCEDF的面积为7225\frac{72}{25}288121\frac{288}{121}
拓展探究:\because正方形ABCDABCD面积为1818
AD=18=32\therefore AD=\sqrt{18}=3\sqrt{2}
AF=yAF=y
SADE=SADF+SAEF\because S_{\triangle ADE}=S_{\triangle ADF}+S_{\triangle AEF}
12ADAB=12AFAD+12AFBE\therefore \frac{1}{2}AD•AB=\frac{1}{2}AF•AD+\frac{1}{2}AF•BE
即:12×32×32=12×32y+12×22y\frac{1}{2}×3\sqrt{2}×3\sqrt{2}=\frac{1}{2}×3\sqrt{2}y+\frac{1}{2}×2\sqrt{2}y
解得:y=925y=\frac{9\sqrt{2}}{5}
AF=925\therefore AF=\frac{9\sqrt{2}}{5}.

解析

延伸应用:
(1)(1)连接CDCD

\because矩形CEDFCEDF是正方形,
DF=DE\therefore DF=DE
ACB=90\because \angle ACB=90^{\circ}BC=3BC=3AC=4AC=4
SABC=12ACBC=12×4×3=6\therefore {S}_{△ABC}=\frac{1}{2}AC•BC=\frac{1}{2}×4×3=6SABC=SBCD+SACD=12BCDF+12ACDE{S}_{△ABC}={S}_{△BCD}+{S}_{△ACD}=\frac{1}{2}BC•DF+\frac{1}{2}AC•DE
12×3DF+12×4DE=6\therefore \frac{1}{2}×3•DF+\frac{1}{2}×4•DE=6,解得:DE=DF=127DE=DF=\frac{12}{7}
\therefore矩形CEDFCEDF的面积=DE2=14449=D{E}^{2}=\frac{144}{49}
故答案为:14449\frac{144}{49}
(2)(2)连接CDCD

\because矩形CEDFCEDF中两边之比为1:21:2
DF=2DEDF=2DE时,设DE=xDE=xDF=2xDF=2x
ACB=90\because \angle ACB=90^{\circ}BC=3BC=3AC=4AC=4
SABC=12ACBC=12×4×3=6\therefore {S}_{△ABC}=\frac{1}{2}AC•BC=\frac{1}{2}×4×3=6SABC=SBCD+SACD=12BCDF+12ACDE{S}_{△ABC}={S}_{△BCD}+{S}_{△ACD}=\frac{1}{2}BC•DF+\frac{1}{2}AC•DE
12×3×2x+12×4x=6\therefore \frac{1}{2}×3×2x+\frac{1}{2}×4x=6,解得:x=65x=\frac{6}{5}
\therefore矩形CEDFCEDF的面积=DEDF=x2x=2x2=2×3625=7225=DE•DF=x•2x=2{x}^{2}=2×\frac{36}{25}=\frac{72}{25}
DE=2DFDE=2DF时,设DE=2aDE=2aDF=aDF=a
ACB=90\because \angle ACB=90^{\circ}BC=3BC=3AC=4AC=4
SABC=12ACBC=12×4×3=6\therefore {S}_{△ABC}=\frac{1}{2}AC•BC=\frac{1}{2}×4×3=6SABC=SBCD+SACD=12BCDF+12ACDE{S}_{△ABC}={S}_{△BCD}+{S}_{△ACD}=\frac{1}{2}BC•DF+\frac{1}{2}AC•DE
12×3a+12×4×2a=6\therefore \frac{1}{2}×3a+\frac{1}{2}×4×2a=6,解得:x=1211x=\frac{12}{11}
\therefore矩形CEDFCEDF的面积=DEDF=2aa=2a2=2×144121=288121=DE•DF=2a•a=2{a}^{2}=2×\frac{144}{121}=\frac{288}{121}
综上,矩形CEDFCEDF的面积为7225\frac{72}{25}288121\frac{288}{121}
拓展探究:\because正方形ABCDABCD面积为1818
AD=18=32\therefore AD=\sqrt{18}=3\sqrt{2}
AF=yAF=y
SADE=SADF+SAEF\because S_{\triangle ADE}=S_{\triangle ADF}+S_{\triangle AEF}
12ADAB=12AFAD+12AFBE\therefore \frac{1}{2}AD•AB=\frac{1}{2}AF•AD+\frac{1}{2}AF•BE
即:12×32×32=12×32y+12×22y\frac{1}{2}×3\sqrt{2}×3\sqrt{2}=\frac{1}{2}×3\sqrt{2}y+\frac{1}{2}×2\sqrt{2}y
解得:y=925y=\frac{9\sqrt{2}}{5}
AF=925\therefore AF=\frac{9\sqrt{2}}{5}.

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