Hello.. and good evening.
I would like to know how to solve for the area of the mid-section of this prismatoid. I tried solving for the volume and I got an answer of 375 cu. ft. but according to Kern and Bland's book the real answer is 310 cu.ft. I guess I made a mistake in solving it.
Can I know the complete solution and the illustration of the top view of the prismatoid's mid-section? Thank you very much to whoever will answer my question. It would be a great help for me.
Problem: Edge AB is perpendicular to the bases BDC and AEF. Find the volume of the solid, if BD is parallel to AE.
Figure: http://imgur.com/7YvdHkS
LI is parallel to GH, thus, GHIL is a trapezoid.
GH = 7/2 = 3.5
HI = 14/2 = 7
JK = 8/2 = 4
IJ = 6/2 = 3
IK = 7/2 = 3.5
KL = 12/2 = 6
LG = 13/2 = 6.5
IL = (7 + 12)/2 = 9.5
IL = IK + LG = 3.5 + 6 = 9.5 (okay)
Area of the mid section = Area of GHIL + Area of IJK
ano tama sagot lumalabas kc sa sagot ko..V=360.59 hehehe
Add new comment