题霸题霸学习平台
← 返回公开题库
八年级数学解答题一般
题目
如图,在同一平面内,直线ll同侧有三个正方形AA,BB,CC,若AA,CC的面积分别为161699,则阴影部分的总面积为____.
知识点:全等三角形的判定、正方形的性质、三角形的面积章节:第21章 四边形 / 21.3 特殊的平行四边形 / 21.3.3 正方形

答案与解析

答案

如图,作LMFELM\bot FEFEFE的延长线于点MM,交JIJI的延长线于点NN
\because四边形AABBCC都是正方形,且正方形AACC的面积分别为161699
EKI=EDR=IHG=90\therefore \angle EKI=\angle EDR=\angle IHG=90^{\circ}DE2=16DE^{2}=16HI2=9HI^{2}=9
DE=4\therefore DE=4HI=3HI=3
EDK=KHI=18090=90\because \angle EDK=\angle KHI=180^{\circ}-90^{\circ}=90^{\circ}
DKE=90KHI=HIK\therefore \angle DKE=90^{\circ}-\angle KHI=\angle HIK
EDK\triangle EDKKHI\triangle KHI中,
{EDK=KHIDKE=HIKEK=KI\left\{\begin{array}{l}{∠EDK=∠KHI}\\{∠DKE=∠HIK}\\{EK=KI}\end{array}\right.
EDK\therefore \triangle EDKKHI(AAS)\triangle KHI\left(AAS\right)
DK=HI=3\therefore DK=HI=3DE=HK=4DE=HK=4
SEDK=SKHI=12×4×3=6\therefore S_{\triangle EDK}=S_{\triangle KHI}=\frac{1}{2}\times 4\times 3=6
DEF=HIJ=90\because \angle DEF=\angle HIJ=90^{\circ}
DEM=180DEF=90\therefore \angle DEM=180^{\circ}-\angle DEF=90^{\circ}HIN=180HIJ=90\angle HIN=180^{\circ}-\angle HIJ=90^{\circ}
KEL=KIL=90\because \angle KEL=\angle KIL=90^{\circ}
MEL=DEK=90KEM\therefore \angle MEL=\angle DEK=90^{\circ}-\angle KEMNIL=HIK=90KIN\angle NIL=\angle HIK=90^{\circ}-\angle KIN
EF\because EFl,IJl,IJll
EF\therefore EFIJIJ
EML=EMN=N=90\therefore \angle EML=\angle EMN=\angle N=90^{\circ}
EML\triangle EMLEDK\triangle EDK中,
{MIL=DEKEML=EDKEL=EK\left\{\begin{array}{l}{∠MIL=∠DEK}\\{∠EML=∠EDK}\\{EL=EK}\end{array}\right.
EML\therefore \triangle EMLEDK(AAS)\triangle EDK\left(AAS\right)
EM=ED=EF\therefore EM=ED=EF
SEFL=SEML=SEDK=6\therefore S_{\triangle EFL}=S_{\triangle EML}=S_{\triangle EDK}=6
LNI\triangle LNIKHI\triangle KHI中,
{NIL=HIKN=KHIIL=IK\left\{\begin{array}{l}{∠NIL=∠HIK}\\{∠N=∠KHI}\\{IL=IK}\end{array}\right.
LNI\therefore \triangle LNIKHI(AAS)\triangle KHI\left(AAS\right)
IN=IE=IJ\because IN=IE=IJ
SLJI=SLNI=SKHI=6\therefore S_{\triangle LJI}=S_{\triangle LNI}=S_{\triangle KHI}=6
SEFL+SLJI=6+6=12\therefore S_{\triangle EFL}+S_{\triangle LJI}=6+6=12
\therefore阴影部分的总面积为1212.

解析

如图,作LMFELM\bot FEFEFE的延长线于点MM,交JIJI的延长线于点NN
\because四边形AABBCC都是正方形,且正方形AACC的面积分别为161699
EKI=EDR=IHG=90\therefore \angle EKI=\angle EDR=\angle IHG=90^{\circ}DE2=16DE^{2}=16HI2=9HI^{2}=9
DE=4\therefore DE=4HI=3HI=3
EDK=KHI=18090=90\because \angle EDK=\angle KHI=180^{\circ}-90^{\circ}=90^{\circ}
DKE=90KHI=HIK\therefore \angle DKE=90^{\circ}-\angle KHI=\angle HIK
EDK\triangle EDKKHI\triangle KHI中,
{EDK=KHIDKE=HIKEK=KI\left\{\begin{array}{l}{∠EDK=∠KHI}\\{∠DKE=∠HIK}\\{EK=KI}\end{array}\right.
EDK\therefore \triangle EDKKHI(AAS)\triangle KHI\left(AAS\right)
DK=HI=3\therefore DK=HI=3DE=HK=4DE=HK=4
SEDK=SKHI=12×4×3=6\therefore S_{\triangle EDK}=S_{\triangle KHI}=\frac{1}{2}\times 4\times 3=6
DEF=HIJ=90\because \angle DEF=\angle HIJ=90^{\circ}
DEM=180DEF=90\therefore \angle DEM=180^{\circ}-\angle DEF=90^{\circ}HIN=180HIJ=90\angle HIN=180^{\circ}-\angle HIJ=90^{\circ}
KEL=KIL=90\because \angle KEL=\angle KIL=90^{\circ}
MEL=DEK=90KEM\therefore \angle MEL=\angle DEK=90^{\circ}-\angle KEMNIL=HIK=90KIN\angle NIL=\angle HIK=90^{\circ}-\angle KIN
EF\because EFl,IJl,IJll
EF\therefore EFIJIJ
EML=EMN=N=90\therefore \angle EML=\angle EMN=\angle N=90^{\circ}
EML\triangle EMLEDK\triangle EDK中,
{MIL=DEKEML=EDKEL=EK\left\{\begin{array}{l}{∠MIL=∠DEK}\\{∠EML=∠EDK}\\{EL=EK}\end{array}\right.
EML\therefore \triangle EMLEDK(AAS)\triangle EDK\left(AAS\right)
EM=ED=EF\therefore EM=ED=EF
SEFL=SEML=SEDK=6\therefore S_{\triangle EFL}=S_{\triangle EML}=S_{\triangle EDK}=6
LNI\triangle LNIKHI\triangle KHI中,
{NIL=HIKN=KHIIL=IK\left\{\begin{array}{l}{∠NIL=∠HIK}\\{∠N=∠KHI}\\{IL=IK}\end{array}\right.
LNI\therefore \triangle LNIKHI(AAS)\triangle KHI\left(AAS\right)
IN=IE=IJ\because IN=IE=IJ
SLJI=SLNI=SKHI=6\therefore S_{\triangle LJI}=S_{\triangle LNI}=S_{\triangle KHI}=6
SEFL+SLJI=6+6=12\therefore S_{\triangle EFL}+S_{\triangle LJI}=6+6=12
\therefore阴影部分的总面积为1212.

AI 自由组卷

围绕这道题再组一份练习 →

完整试卷

浏览同年级试卷结构 →