AB:BC is 3:4. Solve for x.
A.11
B.14
C.120
D.140

- jaylelile

AB:BC is 3:4. Solve for x.
A.11
B.14
C.120
D.140

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- jaylelile

- jaylelile

##### 1 Attachment

- mathstudent55

You are given this, right?
\(\dfrac{AB}{BC} = \dfrac{3}{4} \)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- jaylelile

right..

- mathstudent55

Now replace AB and BC with the lengths given in the figure.
What do you get?

- jaylelile

|dw:1433800191984:dw|

- mathstudent55

\(\dfrac{90}{10x-20} = \dfrac{3}{4}\)
Ok?

- jaylelile

oh okay

- mathstudent55

-20 is not the length of BC. The entire expression 10x - 20 is.
See my equation above?

- mathstudent55

Now we solve that equation.

- mathstudent55

When you have a fraction equal to a fraction, you can cross multiply.

- mathstudent55

|dw:1433800210299:dw|

- jaylelile

so x=14?????

- mathstudent55

|dw:1433800248981:dw|

- jaylelile

x=14 right?

- jaylelile

I just have a hard time making the equations.

- jaylelile

Can you help me with a few more questions?

- mathstudent55

Now we cross multiply in our problem.
\(3(10x - 20) = 4(90) \)
\(30x - 60 = 360\)
\(30x = 420\)
\(x = 14\)
You are correct.

- jaylelile

yay! Thank you so much! Can you help me with a few more? I get confused when it comes to setting up the equations.

- mathstudent55

Let's see if our answer makes sense.
Sides AB and BC are in a 3:4 ratio.
Side AB measures 90.
Side BC measures 10(x) - 20.
We know x = 14, so side BC measure 10(14) - 20 = 140 - 20 = 120.
Side BC measures 120.
Since AB:BC = 3:4, then 90:120 also has to be a 3:4 ratio.
\(\dfrac{90}{120} = \dfrac{30 \times 3}{30 \times 4} = \dfrac{3}{4]\)
We are correct. x = 14 makes sense.

- jaylelile

Find the value of x.
A.
7
B.
9
C.
11
D.
12

##### 1 Attachment

- mathstudent55

Do you know the triangle angle bisector theorem?

- mathstudent55

The angle bisector of an angle of a triangle divides the opposite side in two segments whose lengths are proportional to the lengths of the other sides of the triangle, each segment to its adjacent side.

- jaylelile

I just don't know how to write an equation to solve.

- mathstudent55

That is a very cumbersome way to express this:
|dw:1433800821218:dw|

- mathstudent55

Compare the above figure to your problem.

- mathstudent55

|dw:1433801007502:dw|

- mathstudent55

Here is the fugure from your problem with the letter "y" added.

- mathstudent55

The angle at the right was bisected.
That means look at the segments on the opposite side.
The segments are x and y.
Ok?

- jaylelile

I'm not really following....

- jaylelile

wait! So it would be 9??? I think I'm understanding.

- mathstudent55

According tot he theorem, the segments are proportional to the adjacent sides, so you can write this:
\(\dfrac{36}{x} = \dfrac{28}{y} \)

- jaylelile

so it isn't 9?

- mathstudent55

Now we need to express y in terms of x.
Notice the left side is 16.
One segment in it is x, so the other segment is 16 - x.
That is the segment I called y.
\(\dfrac{36}{x} = \dfrac{28}{16 - x} \)
Now you cross multiply and solve for x.

- mathstudent55

28x = 36(16 - x)
28x = 576 - 36x
64x = 576
x = 9

- mathstudent55

You are correct.

- jaylelile

YAY!!! Thanks! can you help me with one more?

- jaylelile

last one:
Find the value of each variable.
A.
a = 15, b = 5, c = 8, d = 4
B.
a = 15, b = 4, c = 8, d = 5
C.
a = 14.5, b = 5, c = 6, d = 4
D.
a = 14.5, b = 4, c = 6, d = 5

##### 1 Attachment

- mathstudent55

Ok, here there is another thing at work.
If you have two lines that are cut by parallel lines, the segments are of proportional lengths.
That means
\(\dfrac{d}{5} = \dfrac{12}{15} = \dfrac{c}{10} \)

- mathstudent55

Use the left two fractions as a proportion and solve for d.
Then use the right two fractions as a proportion and solve for c

- mathstudent55

Do you understand what to do?

- jaylelile

yes! Thank you very much for all your help!

- mathstudent55

You're welcome.

Looking for something else?

Not the answer you are looking for? Search for more explanations.