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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

AC/DE = AD/DB

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

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.

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.

Re: Proving Triangles Similar! Application problem

Quote:

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

AC/DE = AD/DB

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

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

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

Re: Proving Triangles Similar! Application problem