Question

HELPP!!!

The figure below shows CB = 3, BE = 5, AB = 5x – 2, and DB = 3x + 2

http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_Seg_Two_Exam_36/image0024e9d7cb1.jpg


If ?ABC ~ ?DBE, the value of x is _________

Answer 1

Hi 10jok :)

Sorry your given page is not opening. :(

Answer 2

there

Answer 3

OK. I am not sure. But I want to help you.

There are two triangles. ABC and DBE. If you see both are right triangles. Can you see them?

Answer 4

yea i do

Answer 5

OK. So you know Pythagorean theorem? i.e.\[\Large hyp^2=perp^2+base^2\]

Do you know about this identity?

Answer 6

same as a^2+b^2=c^2 right?

Answer 7

yes yes. right.

in your equation c=hypotenuse, a=perpendicular and b=base. OK

Answer 8

yeaah

Answer 9

OK take a look on both triangles separately.
first ABC
If we apply Pythagorean theorem to this triangle then we can write it as
\[\large hyp^2=perp^2+base^2\]
\[AB^2=AC^2+BC^2\]
so AB=hyp, AC=perp and BC=base

Right?

Answer 10

correct

Answer 11

Similarly for DBE
\[\large DB^2=DE^2+BE^2\]
so DB=hyp, DE=perp and BE=base.

Answer 12

we have to find x tho right not DB?

Answer 13

As in question. the given condition is
ABC is congruent to DBE

This means
these are triangles are equal in magnitude.

So we can write as

\[AB^2=DB^2\]
now put equations of AB and DB
\[AC^2+BC^2=DE^2+BE^2------(1)\]


We will use it later. so that's why I have named it eq(1)

Because we don't have AC and DE

So first we have to find them.

Until now you get it?

Answer 14

yeap!

Answer 15

Ok let us find AC

Use Pythagorean theorem for ABC, which we used above. and separate AC from there. So it look like this.

\[\large AC^2=AB^2-BC^2\]
we have values of AB=5x-2 and BC=3 put them
\[\large AC^2=(5x-2)^2-3^2\]

Now find DE

Use Pythagorean theorem for DBE, which we used above and separate DE from there. So it look like this.

\[\large DE^2=DB^2-BE^2\]
we have values of DB=3x+2 and BE=5 put them
\[\large DE^2=(3x+2)^2-5^2\]

Answer 16

what do we do now ?

Answer 17

Now we can use eq(1) because we have those values.
\[\large eq(1)=>\]
Now put values which we have from diagram and which we have find above

\[\large (5x-2)^2-\cancel3^2+\cancel3^2=(3x+2)^2-\cancel{5^2}+\cancel{5^2}\]
\[\large (5x-2)^2=(3x+2)^2\]
Take square root on both sides.

Now can you solve it further?

Answer 18

wait i could but idk how to take square root from both sides

Answer 19

Basically, you just remove the squares off since both sides have one. It turns to
5x - 2 = 3x + 2
2x = 4
x = 2

Answer 20

\[\large \sqrt{(5x-2)^2}=\sqrt{(3x+2)^2}\]

Answer 21

now square root and square are opposite to each other so they will cancel out.

Answer 22

i thought it was 2 also but if i try to plug it in the actual question it dosent make sense for x to equal 2

Answer 23

so you will left with only \[\large 5x-2=3x+2\]
I hope now you can calculate x.

Answer 24

How are you checking your answer?

Answer 25

in the question, triangle ACB i tried to plug in 2 in 5x-2 and its 8 so i tried to do the pythagorean theorem a^2+3^2=8^2 and didnt come out right

Answer 26

so what is a?

Answer 27

55

Answer 28

sqrt is like 7.5

Answer 29

yes 7.5 approx.

Answer 30

so its 2 ?

Answer 31

yes. it is x=2

Answer 32

ok thank you so much bro you helped me out aloot!! because now i actually know how to do it lol

Answer 33

help me out here whats happening:

or either a rotation 90 degrees counterclockwise about the x axis or y axis?

Answer 34

What to do with this diagram?

Answer 35

there asking me weather its a 90 degrees rotation about the x axis or about the y axis and i think its y but im not 100% sure

Answer 36

It is y-axis rotation.

Answer 37

thanks man take care i will let you know if i need another ques youve been really helpful!

Answer 38

you are ever welcome. :) good luck. just believe on your own self.