Question

The figure shows three right triangles. Triangles ABD, CAD, and CBA are similar.
Theorem: If two triangles are similar, the corresponding sides are in proportion.
Using the given theorem, which two statements help to prove that if segment BC is x, then x2 = 74?

Answer 1

Can we see the figure?

Answer 2

Do you know what we have to do?

Answer 3

Are there any answer choices?

Answer 4

answer choices:
(A) Segment BC ⋅ segment DC = 49
Segment BC ⋅ segment BD = 35

(B) Segment BC ⋅ segment DC = 49
Segment BC ⋅ segment BD = 25

(C) Segment BC ⋅ segment DC = 25
Segment BC ⋅ segment BD = 35

(D) Segment BC ⋅ segment DC = 25
Segment BC ⋅ segment BD = 49

Answer 5

Sorry for not responding, I was afk.

Answer 6

So we could start out by finding out the side lengths of all the sides.

Answer 7

To do that, we can use Pythagorean theorem

Answer 8

Can you find BC?

Answer 9

maybe 7

Answer 10

BC^2=AB^2+AC^2

Answer 11

So BC^2 = 7^2+5^2

Answer 12

So BC = sqrt(49+25)

Answer 13

sqrt74

Answer 14

yes.

Answer 15

So we know the length of that. We can set up a proportion to solve for other lengths.

Answer 16

But sqrt74 is 8.6

Answer 17

hold up. So sqrt(74) is BC Lets find DC and BD

Answer 18

So do you know that we can set up a proportion and solve for similar triangles?

Answer 19

okay. BA/BD = AC/DA = BC/BA.
5/BD = 7/DA = sqrt(74)/5

Answer 20

So to solve for BD
5/BD = sqrt(74)/5
BD(sqrt(74))=25
BD = 25/sqrt(74)

Answer 21

Do you understand how I got that proportion?

Answer 22

yea

Answer 23

Similar triangles. The corresponding sides can be used to set up a proportion.

Answer 24

Okay. now lets find DC.
AC/DC = BA/DA = BC/AC
7/DC = 5/DA = sqrt(74)/7

Answer 25

7/DC = sqrt(74)/7
DC(sqrt(74)) =49
DC = 49/(sqrt(74))

Answer 26

So we have BC = sqrt(74), DC= 49/sqrt(74) and BD = 25/sqrt(74)

Answer 27

Which answer choice matches it now

Answer 28

Oh the answer is (B) Segment BC ⋅ segment DC = 49, Segment BC ⋅ segment BD = 25

Answer 29

On the spot! You got it. Keep it up.

Answer 30

Thank you so much for helping me

Answer 31

np

Answer 32

:) :)

Answer 33

:)

Answer 34

if I have more questions about math, could I ask you for help? :) :) :)

Answer 35

yep, make a new post