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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

First one u just need to use a theoreme of the cosinus in the triangles (sorry for bad london)

4

all I know is 2+2=4-1=3

2 replies

quick math

damn that was fast

How ABC=72° in 2nd lol,I don't understand

8 replies

Angle aBc=72°

7 replies

you need to prove that the angle ABC is 72 degrees right?

4 replies

Yes

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°

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Thx

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2nd angle is <CBO,not <BOC,did a small mistake

2) a) solved come here

i had exam of this 2 days ago but i forgot it all

TFW you realize you have an easier time solving differential equations than basic geometry

1 reply

bump

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

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

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OMG I DIDNT READ THE CONDITIONS CORRECTLY HAHAHAHAHAHAHAHA SORRY MATE

"Top angle is x, therefore bottom 2 angles are 2x. " wtf

if top angle is x that does not mean that angles in the left side and in the right side are 2x

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"<ABC=2<BAC"
<BAC is top angle

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OMFG i didnt read correctly conditions haha

I thought that angle ABC is 2 angles DBC (and that was obvious cuz BD is isossless

Thanks a lot
Can you explain a bit more on 1. b) please?

1 reply