parallelepiped

Forums: Science And Math
Email this Topic • Print this Page

Hello!

Here is web address where you can found a real picture and calculated problem

http://matematika.agava.ru/S_1951.html

My problem is (only) in second formula .
VBAEC=1/3*hB*SAEC=1/3*1*(1/2*EO*sin60*AC)=1/3*1/2*R*sqrt(3)/2*2*R/sqrt(2)=R^2*sqrt(6)/12

How is in area of triangle this possible( 1/2*EO*sin60*AC) IF YOU BE SO KIND TO LOOK AT THE PICTURE ON WEB ADDRESS I` GAVE TO YOU .

PLEASE HELP!!!

Topic Stats
Top Replies
Link to this Topic

Type: Discussion • Score: 1 • Views: 535 • Replies: 4

No top replies

EO * sin60 is the altitude of AEC where AC is the base.

It follows that the area of AEC is 1/2 * (EO * sin60) * AC.

0 Replies

Hello!

Thank you!!
But can I` ask(stupid question) ,
for example if there is angle of 90 instead of 60 then we would write :
EO*sin90*AC since sin90=1 then we do not write it and that is why we usualy write EO*AC/2 , (S= h*a/2) . is this near the thrue.

Thanks so much!!

parallelepiped

Related Topics

Quick Links

My Account

able2know