Question

In △ABC, BD is an altitude.
Image posted. What is the length, in units, of BD?
5, 3, 6 radical 5, or 5 radical 6.

Answer 1

BC^2 = CD * DA.
Find BC, then use pythagoras and there you go.

Answer 2

Would that be 5?

Answer 3

Uhm, I am not sure. BC = \[5\sqrt6\] and it is the hypotenuse of the triangle BCD. Now..
BC^2 = CD^2 + BD^2 =>
BD^2 = BC^2 - CD^2.
BD^2 = 150 - 100
BD ^2 = 50;
BD = \[\sqrt50\]
BD = \[5\sqrt2\]

Answer 4

Ok then that would give me my final answer.

Answer 5

I'm guessing it would turn out to be 3.

Answer 6

The answer to 5 radical 2 is 7.

Answer 7

Woops, my bad, the answer is that one, the last one, but BC^2 = 25*10 so BC = 5radical10 , now follow up the calcs and you'll turn out to be right. I misread the letters

Answer 8

\[(\sqrt250)^2 - 100 = BD^2\] which is \[5\sqrt6\]

Answer 9

Ok so now I have to do 5 radical 10 and then that will give me my answer?

Answer 10

oh ok 5 radical 6.

Answer 11

Now I understand that question. Thanks!

Answer 12

No problem, I kinda mis-read the letters. Anyway these are the properties you want to learn:

The square of the length of a cathetus equals the product of the lengths of its orthographic projection on the hypotenuse times the length of this.

b² = a · m
c² = a · n

Answer 13

I'll try to remember those! ;)

Answer 14

By the way, if you think about it, with the very first formula I used, it worked, so you could have used that one from the very beginning without doing pythagoras ! But anyhow, more math can't be bad. :)

Answer 15

Thanks for your help! I actually have 2 more like that I think. :/