Question #672c6

1 Answer
Aug 19, 2017

drawn

As stated in the problem pyramid V-ABCD 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 #Delta VDC=sqrt(24(24-18)(24-18)(24-12))"in"^2#

#=sqrt(24xx6xx6xx12)"in"^2=72sqrt2"in"^2#

  • 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#