Question #672c6
1 Answer
As stated in the problem pyramid VABCD is cut from a cube of edge 12 in.,as shown in the figure. The vertex V is the midpoint of an upper edge of the cube. We are to compute the lateral surface of the pyramid.

The lateral surface of the pyramid composed of 4 triangles
#Delta VAB,Delta VDC , DeltaVBC=DeltaVAD# 
Area of
#Delta VAB=1/2ABxxVE=1/2xx12xx12"in"^2=72"in"^2# 
Length of
#VB=sqrt(EV^2+EB^2)=sqrt(12^2+6^2)=6sqrt5"in"# 
In
#DeltaVBC,/_VBC=90^@# 
Length of
#VC=sqrt(BV^2+BC^2)=sqrt((6sqrt5)^2+12^2)=18"in"# 
Area of
#Delta VBC=1/2BCxxVB=1/2xx12xx6sqrt5=36sqrt5"in"^2# 
So Area of
#Delta VAD=36sqrt5"in"^2# 
Now semi perimeter of of
#Delta VDC=1/2(18+18+12)=24"in"#
*So area of
 Hence the lateral surface of the pyramid composed of 4 triangles
#=Delta VAB+Delta VDC + DeltaVBC+DeltaVAD#
#=72+72sqrt2 + 36sqrt5+36sqrt5=72(1+sqrt2+sqrt5)"in"^2#