Question

help please

Answer 1

@563blackghost

Answer 2

Due to the fact this is a square the diagonals of the square are all congruent.

GL equals \(\large\bf{\sqrt{2}}\) since the diagonals are equal then `HL is equal as well` we would add `GL's length twice to get HK`.

\(\huge\bf{\sqrt{2} + \sqrt{2} =?}\)

Answer 3

it would be 2.82 if you round then 3 @563blackghost

Answer 4

Correct :) Though it be best to keep format so I would believe you would put \(\large\bf{2\sqrt{2}}\).

Now due to the fact that this is a square the angles of the diagonals are equal to 90 degrees. So `KLJ` equals 90 degrees.

Answer 5

ooo okay

Answer 6

what about HK? @563blackghost

Answer 7

\(\large\bf{2\sqrt{2}}\) was HK. Sorry if that was not clear.

Answer 8

oh okay thank you

Answer 9

i probably just didn't read it right

Answer 10

Since the transversals of the square bisect the angle of `J` that would mean that the angles on either side are equal to each other. Due to this knowledge we divide `90` by 2 to find `Angle HJG`

\(\huge\bf{\frac{90}{2 }=\angle HJG}\)

Answer 11

can you explain that in an simpler form ^.^ if not its okay

Answer 12

an angle bisector bisects the angle equally so that on either side they have the same degree. This happens when there are diagonals in a square. `Angle J` is bisected by a diagonal giving `angle HJG and angle KJG` since it bisects it the two made angles are congruent. So we would divide the total angle degree `(90)` by 2 to get `angle HJG`

Answer 13

okay makes more sense

Answer 14

So `HJG would equal 45 degrees`

Answer 15

4. Due to the fact we found the length of the hypotenuse of the `triangle HJK` being \(\large\bf{2\sqrt{2}}\) and we have the side lengths `HJ and KJ` of both being two `(this is because it is a square having equal side lengths)` We would add them up to find the perimeter.

\(\huge\bf{2 \sqrt{2} + 2+2=perimeter}\)

Answer 16

and that would be 6.66 i think

Answer 17

I got `6.828...` you would round it to `6.8`

Answer 18

ohh ok oops

Answer 19

All's good ^.^

Now to the next

Answer 20

In a rhombus its opposite sides are congruent to each other. And its opposite angles are congruent to each other as well. Due to the fact it has diagonals in the rhombus, they each would be equal to each other. With that said which would apply to always being true?

Answer 21

in a rhombus its opposite sides are congruent to each other. And its opposite angles are congruent to each other as well. that would be true

Answer 22

Yes quite :)

So that would mean that \(\large\bf{\angle A \cong \angle C}\) and \(\large\bf{AC \cong BD}\).

Answer 23

wow your so fast at solving these

Answer 24

Now I do believe the bisecting of angles includes for rhombus meaning that \(\large\bf{\angle ABD \cong \angle CBD}\)

Answer 25

so that would apply to?

Answer 26

I believe so, yes.

Answer 27

ohh okay thank you so much

Answer 28

No problem Alexis :)