Hello, I need some help. I got the answer 9, but it should be 4,5. What am I doing wrong?
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Question

anonymous · Answer

I don't see anything here

kamille · Answer

attachment

kamille · Answer

Well, I got a=3, because V=a^3 then 
So right now we are looking at triangle ADC:
$$AC ^{2}=3^{2}+3^2=18$$
$$AC=\sqrt{18}=3\sqrt{2}$$

then I need half of AC(lets call it AO) is $$AO=\frac{ 3\sqrt{2} }{ 2 }$$

heighten of triange ADC (lets call it DO) is
$$DO ^{2}=3^{2} - (\frac{ 3\sqrt{2} }{ 2 })^{2}=9-4.5=4.5$$
$$DO=\sqrt{4,5}=\sqrt{\frac{ 9 }{2 }}=3\sqrt{\frac{ 1 }{2 }}$$

So S of triangle ADC is S=DO*AC=9

V=(1/3)*9*3=9 BUT THE ANSWER SAYS IT SHOULD BE 4,5

anonymous · Answer

can you tell me the whole problem as well?

kamille · Answer

ahh, sorry! I need to find V of a pyramid!

kamille · Answer

butttt also, I have found my mistake S of a triangle is S=(ah)/2

anonymous · Answer

answer was 4,5? does that mean half of 9?

kamille · Answer

yup

anonymous · Answer

so you only need to substitute the S=ah with S=ah/2, which give you S=DO*AC/2=9/2

anonymous · Answer

do you get it?