Question

help


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Answer 1

how to find the second one?

Answer 2

The diagonals are in the middle of the triangle so its basically it is breaking it up in 4 equal peices.

Answer 3

pok

Answer 4

ok

Answer 5

then

Answer 6

So if you are trying to find ac it would what half of it twice.

Answer 7

so (x+4)(x+4)

Answer 8

simplify?

Answer 9

Since you have the sum of the length of the diaganols you have to make an equation equal to 20.

Answer 10

(2x-3)(2x-3)+(x+4)(x+4)=20

Answer 11

You have to solve for x before moving on

Answer 12

is the answer 7?

Answer 13

Uh one sec still solving!

Answer 14

Plug it in to check!

Answer 15

I dont think thats it?

Answer 16

idk

Answer 17

i solve this for x (2x-3)(2x-3)+(x+4)(x+4)=20

Answer 18

You have to do the quadratic eqaution...

Answer 19

idk how to do quad

Answer 20

did you get 5x^2-4x+25=20 when you simplified?

Answer 21

yep

Answer 22

Okay you need quad to find x give me a minute and ill do it...

Answer 23

ok

Answer 24

Its an imaginary number...

Answer 25

Assuming the rhombus is ABCD and E is the intersection of the diagonals, then we know the diagonals perpendicularly bisect each other, so AE is half of one diagonal and BE is half the other diagonal. That allows us to solve for x:
(x+4) + (2x-3) = 20/2
3x + 1 = 10
3x = 9
x = 3
and since
BE = 2x - 3 = 3 is half of BD,
BD = 2*3 = 6 (answer C).

Answer 26

Figured out the better way to solve it lol

Answer 27

thank you

Answer 28

DO you understand how i got there?

Answer 29

so BE is half and BD is the other half?

Answer 30

each is 3?

Answer 31

Wait!!! i messed up lol you plug the three in x+4 and multiply it by 2 because you are finding ac

Answer 32

For some reason i thought it was the other one. But it will be 14 in that case.

Answer 33

thanks