如图∠bca=90°pc⊥面abc则在三角形abc三角形pac的边所在的直线中 1与pc垂直的直线有acabbc 2与 Sketch Coloring Page

如图∠bca=90°pc⊥面abc则在三角形abc三角形pac的边所在的直线中 1与pc垂直的直线有acabbc 2与 Template

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Take a peek at some of the sketches created by our users, are you a sketchite?

sketch #3459 BLUE from fosters imagenary friends

anonymous

anonymous
sketch #3459 BLUE from fosters imagenary friends

sketch #5123

anonymous

anonymous
sketch #5123

sketch #5094 Squirrel by Mehdi Público

anonymous

anonymous
sketch #5094 Squirrel by Mehdi Público

sketch #3693 Lucy from Elfen Lied by Simona Honigmelone

anonymous

anonymous
sketch #3693 Lucy from Elfen Lied by Simona Honigmelone

sketch #4209 bushy

anonymous

anonymous
sketch #4209 bushy

sketch #1169 ENJOY IT!!!!!!!! OH NO EAT IT OR IT WILL MELT!!!!! :0

anonymous

anonymous
sketch #1169 ENJOY IT!!!!!!!! OH NO EAT IT OR IT WILL MELT!!!!! :0

sketch #5263

anonymous

anonymous
sketch #5263

sketch #664 Spider on a plant pot

Rachel Roe
sketch #664 Spider on a plant pot

sketch #3000

anonymous

anonymous
sketch #3000

sketch #3590

anonymous

anonymous
sketch #3590

sketch #5231

anonymous

anonymous
sketch #5231

sketch #978

anonymous

anonymous
sketch #978

sketch #5326

anonymous

anonymous
sketch #5326

sketch #245 Fishy

Lucy Barnard
sketch #245 Fishy

sketch #2888 Jimi Hendrix by Meen Suksoi

anonymous

anonymous
sketch #2888 Jimi Hendrix by Meen Suksoi

sketch #3659 Portrait by Mari Falcsik

anonymous

anonymous
sketch #3659 Portrait by Mari Falcsik

sketch #5257

anonymous

anonymous
sketch #5257

sketch #33 Flutter-bys

Rachel Roe
sketch #33 Flutter-bys

sketch #1703

Laura Agapi Asnicar
sketch #1703

sketch #3701 amazing Skull

anonymous

anonymous
sketch #3701 amazing Skull

sketch #778

Benjo Arellano
sketch #778

sketch #5070 The Rolling Stones by Roby Krdoso

anonymous

anonymous
sketch #5070 The Rolling Stones by Roby Krdoso

sketch #4965

anonymous

anonymous
sketch #4965

sketch #5206

anonymous

anonymous
sketch #5206

sketch #3857

anonymous

anonymous
sketch #3857

sketch #323

anonymous

anonymous
sketch #323

sketch #819

Laura Agapi Asnicar
sketch #819

sketch #2640 Marge by Mirnes Sinanovic

Mirnes Sinanovic
sketch #2640 Marge by Mirnes Sinanovic

sketch #5079 SketchToy Pencil by Sarah Prieto

anonymous

anonymous
sketch #5079 SketchToy Pencil by Sarah Prieto

sketch #76

Rachel Roe
sketch #76

tomski如图∠bca=90°pc⊥面abc则在三角形abc三角形pac的边所在的直线中 1与pc垂直的直线有acabbc 2与 coloring

如图∠BCA=90°PC⊥面ABC则在三角形ABC三角形PAC的边所在的直线中 1与PC垂直的直线有ACABBC 2与 sketch template

如图∠BCA=90°PC⊥面ABC则在三角形ABC三角形PAC的边所在的直线中 1与PC垂直的直线有ACABBC 2与 sketch template

如图在四边形ABCD中∠ABC=∠ADC=90°AB=BC=10AD=2DC连对角线ACBD相交于点E1�求证BD平分∠ sketch template

如图在 ABC中∠ACB=90°CA=CB点P在 ABC内且PA=3PB=1PC=2求 CPB的度数．小东同学尝试在CP的右侧 sketch template

$如图四棱锥P ABCD中侧面PAD为等边三角形且垂直于底面ABCDAB=BC=$frac{1}{2}$AD∠BAD=∠ABC=90 sketch template$

如图，在三棱锥P ABC中，平面PAC⊥ 平面ABC，∠ PCA=90°， ABC是边长为4的正三角形，PC=3，M是AB边上的一动点，则PM sketch template

如图在三棱锥P ABC中∠APB=90°∠PAB=60°AB=BC=CA平面PAB⊥平面ABC．Ⅰ求直线PC与平面ABC所成角的 sketch template

18．如图所示在四棱锥P ABCD中底面ABCD是一直角梯形∠BAD=90° AD∥BC AB = BC A AD 2a sketch template

如图在直角三角形ABC中∠BAC=90°将 ABC沿直线BC�右平移得到 DEF连结ADAE则下列结论中不成立的是 A．AD∥ sketch template

如图在Rt ABC中∠BAC=90°AB=5AC=12P为边BC上一动点PE⊥AB于EPF⊥AC于FM为EF中点则AM的取值 sketch template

如图在三棱锥P ABC中AC=BC=2∠ACB=90°AP=BP=ABPC⊥AC．Ⅰ求证PC⊥ABⅡ求二面角B AP C sketch template

如图 ABC中∠ACB=90°AC=6cmBC=8cm．点P从A点出�沿A C路径�终点C运动点Q从B点出�沿B C A路径�终点A sketch template

如图在 ABC中∠ABC=90°AB=6BC=8点MN分别在边ABBC上沿直线MN将 ABC折叠点B落在点P处如果AP∥ sketch template

题目探究如图� 在四边形ABCD中∠BAD=∠BCD=90°AB=ADAE⊥CD于点E．若AE=10求四边形ABCD的面积 sketch template

17． 如图所示在四棱锥P ABCD中∠ABC＝∠ACD＝90°∠BAC＝∠CAD＝60°PA⊥平面ABCDE为PD的中点PA sketch template

如图所示在直角梯形BCEF中∠CBF=∠BCE=90°AD分别是BFCE上的点AD∥BC且AB=DE=2BC=2AF．将四边形 sketch template

如图在四棱锥P ABCD中PA⊥平面ABCD四边形ABCD为直角梯形AD//BC且AD&gtBC∠DAB=∠ABC=90°PA=AB sketch template

如图在半径为2的扇形AOB中∠AOB=90°点C是弧AB上的一个动点OD⊥BCOE⊥AC垂足分别为DE．1求线段DE的长 sketch template

如图，在三棱锥P ABC中，AB=BC=2√ 2，PA=PB=PC=AC=4，O为AC的中点（1）证明：PO⊥ 平面ABC；（2）若点M在棱 sketch template

18．如图已知斜三棱柱ABC A1B1C1中∠BCA=90°AC=BC=a点A1在底面ABC上的射影恰为AC的中点DBA1⊥AC1 sketch template

如图所示， ABC是等腰直角三角形，∠ACB=90°，AD是BC边上的中线，过C作AD的垂线，交AB于点E，交AD于点F．求证：（1）∠ADC sketch template

三棱锥P ABC中∠ABC=90°PA=PB=PC则下列说法正确的是 A平面PAC⊥平面ABCB平面PAB⊥平面PBCCPB⊥ sketch template

在如图所示的几何体中四边形ABCD为矩形平面ABEF⊥平面ABCDEF∥AB∠BAF=90°AD=2AB=AF=2EF=1点P在 sketch template

17． 如图所示在四棱锥P ABCD中∠ABC＝∠ACD＝90°∠BAC＝∠CAD＝60°PA⊥平面ABCDE为PD的中点PA sketch template

如图点D在BC的延长线上DE⊥AB于E交于AC于F∠B＝50°∠CFD＝60°则∠ACB＝．题目和参考答案——青夏教育精英家教�—— sketch template

在如图所示的几何体中四边形ABCD为矩形平面ABEF⊥平面ABCDEF∥AB∠BAF=90°AD=2AB=AF=2EF=1点P在 sketch template

如图在四边形ABCD中AB=AD∠BAD=∠BCD=90°连接AC若AC=6则四边形ABCD的面积为解析如图作AM⊥BC sketch template

如图在半径为2的扇形AOB中∠AOB=90°点C是弧AB上的一个动点OD⊥BCOE⊥AC垂足分别为DE．1求线段DE的长 sketch template

18．如图已知斜三棱柱ABC A1B1C1中∠BCA=90°AC=BC=a点A1在底面ABC上的射影恰为AC的中点DBA1⊥AC1 sketch template

如图在边长为2的等边 ABC中AD⊥BC于点D且AD=$sqrt{3}$E为AC中点P为AD上一点则 PEC�长的最小值是 sketch template

如图在四边形ABCD中∠A=60°∠B=∠D=90°BC=2CD=3则AB= 5 A C D ——青夏教育精英家教�—— sketch template

如图在平行四边形ABCD中∠DBC=45°DE⊥BC于EBF⊥CD于FDEBF相交于HBFAD的延长线交于P．下面结论� sketch template

如图在三棱锥PABC中AC = BC 2∠ACB 90°AP BP ABPC⊥AC． 1 求证PC⊥AB sketch template

如图，在四棱锥P—ABCD中，平面PAD⊥平面ABCD ， AD//BC AB＝BC＝PA＝1，AD＝2，∠PAD＝∠DAB＝90°，点E sketch template

8．如图在⊿ABC中AB⊥ACAD⊥BCD是垂足则类似有�题“三棱锥A BCD图2中AD⊥平面ABCAO⊥平面BCDO sketch template

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