In the diagram below, BC is an altitude of ABD. To the nearest whole unit, what is the length of CD? (Can anyone solve this)

Question

anonymous · Answer

attachment

e.mccormick · Answer

I would start by solving for AB to get more information.

e.mccormick · Answer

Is this trig or geometry.  If you know the trig ratios, like law of sines, then once you have AB, solving the rest of the triangle is not too hard.  That would get you AD and AD-AC=CD.

anonymous · Answer

this is geomertry );

rajee_sam · Answer

it is repeated use of Pythagorean theorem

e.mccormick · Answer

Yah, that is the geometry method.

e.mccormick · Answer

You have three right triangles there.  With the 2 sides, you can find one 3rd side.  Then it is representing things in terms of what you know until it is solveable.

e.mccormick · Answer

$16^2+30^2=AB^2$
$30^2+CD^2=BD^2$
$AB^2+BD^2=(16+CD)^2$

If you use what is known in the first two to rewrite the third, you can make a formula with only yhr CD as an unknown.

anonymous · Answer

okay ill try tht right now and let you kno wht i get

e.mccormick · Answer

Do you see how each of those is a Pythagorean Theorem appleid to one of the triangles?

anonymous · Answer

okay idnt know what doing

e.mccormick · Answer

$\triangle\mathrm{ABC}\implies AC^2+BC^2=AB^2\implies 16^2+30^2=AB^2$

e.mccormick · Answer

Now do you see what I was looking at?

anonymous · Answer

okay i get thats what it stands for
for 16sqrt i get 256 thn for 30sqrt i get 900 then i add them up and get 1156

anonymous · Answer

then idk what to do from there

anonymous · Answer

then idk what to do from there

e.mccormick · Answer

hmm...  I don't know why you would get a square root in there.  Can you type in what you used for a subsititution?

e.mccormick · Answer

In $AB^2+BD^2=(16+CD)^2$ you need to change $AB^2$ and $BD^2$ into either numbers or CD.  Fortunatly, you have the other triangles, which have $16^2+30^2=AB^2$ and $30^2+CD^2=BD^2$.  Because $AB^2$ and $BD^2$  are squared in all the equations, you can do a direct substitution.  No need to do any roots.

e.mccormick · Answer

Once you have that, show me.

e.mccormick · Answer

Looks like you closed it, just so you see the way I solved it:

$AC^2+BC^2=AB^2$ and
$BC^2+CD^2=BD^2$ and
$AB^2+BD^2=AD^2\implies $
$16^2+30^2=AB^2$ and
$30^2+CD^2=BD^2$ and
$AB^2+BD^2=(16+CD)^2\implies $
$16^2+30^2+30^2+CD^2=(16+CD)^2\implies$
$256+900+900+CD^2=256+32CD+CD^2\implies$
$256-256+1800+CD^2-CD^2=32CD\implies$
$1800=32CD\implies$
$1800/32=CD\implies$
$56.25=CD$

anonymous · Answer

Answer is 56