Question

Question #672c6

Answer 1

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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`