# Thread: Proving Triangles Similar! Application problem

1. ## Proving Triangles Similar! Application problem

Hey Guys,

For #25)

I am dealing with proportionals here with this problem and I don't know how to get the answer to this one.

I tried 10/8 = 4/x

I ended up getting x = 3.2 but the answer is 16 .

I have an idea that will get me 16 but... I honestly don't know if it's right

so I set up the equation as 10/4 = x/8 which gets me x =20
then I plug in 20 into x because DB = AB - AD and I get 20-4 = 16

But if I plug in numbers like this, I don't understand why I should set it up as AC / AD = DE / DB.

2. ## Re: Proving Triangles Similar! Application problem

Hey Cake.

With similarity you have AB/DB = AC/DE but AB = AD + DB which means

(AD + DB)/DB = AC/DE which implies
AD + DB = AC*DB/DE which imlies
DB(AC/DE - 1) = AD which implies
DB = AD / [AC/DE - 1] if my working out is correct.

3. ## Re: Proving Triangles Similar! Application problem

Originally Posted by Cake
Hey Guys,

For #25)

I am dealing with proportionals here with this problem and I don't know how to get the answer to this one.

I tried 10/8 = 4/x

I ended up getting x = 3.2 but the answer is 16 .

I have an idea that will get me 16 but... I honestly don't know if it's right

so I set up the equation as 10/4 = x/8 which gets me x =20
then I plug in 20 into x because DB = AB - AD and I get 20-4 = 16

But if I plug in numbers like this, I don't understand why I should set it up as AC / AD = DE / DB.

Let DB =x
x/x+4 = 8/10

4. ## Re: Proving Triangles Similar! Application problem

Triangles ABC and ADE are similar thus we have
BD/AB = DE/AC
BD/( BD + AD) = 8/10
10 * BD = ( BD + 4) * 8
10 BD = 8 BD + 32
That gives BD = 16

5. ## Re: Proving Triangles Similar! Application problem

Thank you guys!