# Acute Angled Triangle...difference of two sides

• Nov 19th 2011, 03:42 AM
cosmonavt
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

?
• Nov 19th 2011, 05:13 AM
cosmonavt
Re: Acute Angled Triangle...difference of two sides
OK guys I got it!

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.
• Nov 19th 2011, 05:37 AM
earboth
Re: Acute Angled Triangle...difference of two sides
Quote:

Originally Posted by cosmonavt
OK guys I got it!

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:

$(x-1)^2 - (x-k)^2 = (x+1)^2-k^2$

which simplifies to

$2kx-2x=x^2+2x~\implies~k=\frac12 x +2$

3. Now determine using the result from #2

$k - (x-k)$

which yields 4.
• Nov 19th 2011, 06:26 AM
cosmonavt
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.