Question

Non-circle question : Polygons

Answer 1

We can find that hyp and we know an angle of that triangle

Answer 2

\[a^2+a^2=hyp^2 => hyp=\sqrt{2a^2}=a \sqrt{2}\]
what
nevermind
this is all i have lol

Answer 3

I got this, but don't know how to continue. That's the problem

Answer 4

BCE right angle?

Answer 5

Not given

Answer 6

no. don't think so. as well as not given in question too.

Answer 7

I think it is E

Answer 8

not sure.

Answer 9

hard lol

Answer 10

I agree!!!

Answer 11

i need some assumptions here ^^

Answer 12

as it is rhombus then\[\angle BDE=45 ^{o}\]

Answer 13

how come 45?

Answer 14

@Shayaan_Mustafa not possible, angle BDC = 45

Answer 15

its impossible lol .. ive seen

Answer 16

@Callisto yes i know BDC=45
but in rhombus two angles are equal and other two so.
so i took it as..... lol :s

Answer 17

BF parallel to DE, thats why

Answer 18

impossible BDE 45

Answer 19

BDC=45 , BDE<BDC
=> BDE <45

Answer 20

BCE = 45 degress ..

Answer 21

so \[p ^{2}+q ^{2}=4\sqrt{2}a\]

Answer 22

can you see how?

Answer 23

i mean \[x ^{2}+y ^{2}=4\sqrt{?}\]

Answer 24

i posted my calculation above. lol. sorry for that . that are x and y not p and q.

Answer 25

4sqrt(2)a

Answer 26

where is the right angle?? what thm are you using?