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Geometry help
SA | 
Sweden ve! 
1)In the next figure is represented the rhombus ABCD and square ABEF, which have the common side AB. It is known that AB = 6 cm and AC = 6 radical3 cm. imgur.com/a/sePDkOq a)show that ABC=120° b)calculate the area of Triangle BCE 2)The next figure shows an isosceles triangle ABC, AB = AC, BC = 8 cm and <ABC = 2 <BAC. BD segment is the bisector of the angle ABC, D € AC imgur.com/a/V8EHDrS a)show that ABC=72° b)calculate the lenght of AD 3)In the next figure is represented a rectangular parallelepiped ABCDEFGH, and M and N are the middle of the edges AB and CD. It is known that AD = 5 cm and AB = 2AD imgur.com/a/sXm7Afm a)Prove that the plans (DMH) and (BNF) are parallel. b)calculate the distance between the two plans (DMH) and (BNF) Sorry if bad translation,if you cant help please bump Preferably if you solve 1 problem at a time Any help is appreciated
2021-06-13 12:24
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First one u just need to use a theoreme of the cosinus in the triangles (sorry for bad london)
2021-06-13 12:25
#3
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Europe onepoint6
4
2021-06-13 12:26
all I know is 2+2=4-1=3
2021-06-13 12:26
2 replies
#5
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Czech Republic Hap0vecXD
quick math
2021-06-13 12:29
damn that was fast
2021-06-13 13:15
How ABC=72° in 2nd lol,I don't understand
2021-06-13 12:33
8 replies
#8
SA | 
Sweden ve!
Angle aBc=72°
2021-06-13 12:35
7 replies
AB=AC=> <ABC=<ACB.In triangle <ABC+<BAC+<ACB=180°.We change <BAC to 1/2<ABC and <ACB to <ABC,so we get 2,5<ABC=180°=> <ABC=72°
2021-06-13 12:41
you need to prove that the angle ABC is 72 degrees right?
2021-06-13 12:42
4 replies
#13
SA | 
Sweden ve!
Yes
2021-06-13 12:42
3 replies
1 a) In rhombus AC _|_ BD,So we get triangle AOB(O-point between AC and BD,<O=90°) AO=OC=3radical3 Sin<ABO=AO/AB=3radical3/6=radical3/2=> <ABO=60°.Same for triangle CBO.<ABO+<CBO=2<ABO=60*2=120°
2021-06-13 12:59
2 replies
#16
SA | 
Sweden ve!
Thx
2021-06-13 12:55
1 reply
2nd angle is <CBO,not <BOC,did a small mistake
2021-06-13 13:00
2) a) solved come here
2021-06-13 12:42
#7
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Slovenia dootie
i had exam of this 2 days ago but i forgot it all
2021-06-13 12:34
TFW you realize you have an easier time solving differential equations than basic geometry
2021-06-13 12:39
1 reply
#29
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Indonesia segopecel
school/uni put more emphasis on analytical math bcs we are depending more and more on computer(at least in engineering), thus make us more familiar in rows of equations than a genuine pen and paper solution(or maybe chalk and board)
2021-06-13 13:19
bump
2021-06-13 12:39
1. ABD and BCD are equilateral triangles therefore angles ABD and DBC are both 60 60 + 60 = 120 You can work out triangle BCE by using angle BCE = 360-120-90, and BC and BE are the same 2. Top angle is x, therefore bottom 2 angles are 2x. Therefore total value is 5x. x = 36 therefore angle ABC is 72 if it bisects angle 2x, you get angle DBC = x, therefore the 2 triangles are similar. Just find the ratio
2021-06-13 12:58
11 replies
3. Show the symmetry of the shape: literally the same thing on either side. AM = AD, so both planes are at an angle of 45 degrees. therefore, the line NA will be at a normal to DM at it's halfway point. Using pythagoras, you get NA = sqrt50, or 5sqrt2. Divide that by 2 to get 5/2sqrt2
2021-06-13 13:02
1 reply
OMG I DIDNT READ THE CONDITIONS CORRECTLY HAHAHAHAHAHAHAHA SORRY MATE
2021-06-13 13:10
"Top angle is x, therefore bottom 2 angles are 2x. " wtf
2021-06-13 13:05
if top angle is x that does not mean that angles in the left side and in the right side are 2x
2021-06-13 13:06
4 replies
"<ABC=2<BAC" <BAC is top angle
2021-06-13 13:08
3 replies
OMFG i didnt read correctly conditions haha
2021-06-13 13:11
I thought that angle ABC is 2 angles DBC (and that was obvious cuz BD is isossless
2021-06-13 13:12
gyazo.com/0195955650c2847d705e7f3e04a335.. gyazo.com/265c730a477428e1c091469d33e3df.. I was literally solving this sh*t for 20 minutes and then I found out that the conditions are much easier hahahahahahah And I found out that the Angle B is approximately 76 degrees
2021-06-13 13:13
#28
SA | 
Sweden ve!
Thanks a lot Can you explain a bit more on 1. b) please?
2021-06-13 13:19
1 reply
Essentially, I didn't explain it much, but you have 2 edges and an angle, which is enough to work out the triangle. BC and BE are given/deduced, and angle CBE is 150 degrees. Using the cosine rule, sine rule and a bit of pythagoras, you can work it out. I'll try to make you work it out by yourself, but I can do the thing for you if you want between 2 matches I'm refereeing
2021-06-13 14:21
#30
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Finland cuntycorn
"1. ABD and BCD are equilateral triangles therefore angles ABD and DBC are both 60 60 + 60 = 120" I think you have to prove they are equilateral first
2021-06-13 13:40
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