1 Attachment(s)

Acute Angled Triangle...difference of two sides

I am totally clueless on this. I have tried Pythagoras theorem and tried taking a lot of ratios but don't seem to get anywhere:

The length of the sides of the acute angled triangle ABC are x-1, x and x+1. BD is perpendicular to AC. Then CD-DA equals

(a) x/8

(b) x/9

(c) 2

(d) 4

?

Re: Acute Angled Triangle...difference of two sides

OK guys I got it!

The answer is four.

First we equate

cos C = CD/(x+1) and cos C = [ (x+1)^2 + x^2 - (x-1)^2 ] / [2(x+1)(x)]

to find an expression for CD in terms of x then

cos A = AD/(x-1) and cos A = [ (x-1)^2 + x^2 - (x+1)^2 ] / [2(x-1)(x)]

to find an expression for DA in terms of x. Then subtracting CD-DA gives 4.

1 Attachment(s)

Re: Acute Angled Triangle...difference of two sides

Quote:

Originally Posted by

**cosmonavt** OK guys I got it!

The answer is four.

First we equate

cos C = CD/(x+1) and cos C = [ (x+1)^2 + x^2 - (x-1)^2 ] / [2(x+1)(x)]

to find an expression for CD in terms of x then

cos A = AD/(x-1) and cos A = [ (x-1)^2 + x^2 - (x+1)^2 ] / [2(x-1)(x)]

to find an expression for DA in terms of x. Then subtracting CD-DA gives 4. **<--- correct! (Clapping)**

Here is a slightly different approach:

1. I've modified your sketch a little bit (see attachment)

2. Using Pythagorean theorem you'll get:

which simplifies to

3. Now determine using the result from #2

which yields 4.

Re: Acute Angled Triangle...difference of two sides

Yeah! That's way more simpler. Wonder why I didn't get that in the first place.