SOLUTION: for homework there is a challenge question that says: This net will fold into a triangular pyramid. Find the surface area of the pyramid, and estimate its volume. The heigh

Question 62559: for homework there is a challenge question that says:
This net will fold into a triangular pyramid. Find the surface area of the pyramid, and estimate its volume.

The height of the pyramid is 5.2cm and the length is 6cm
I got this from the Math Impact Textbook, course 2(purple book)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
for homework there is a challenge question that says:
This net will fold into a triangular pyramid. Find the surface area of the pyramid, and estimate its volume.
The height of the pyramid is 5.2cm and the length is 6cm
I got this from the Math Impact Textbook, course 2(purple book)
base of triangular pyramid given is an equilateral triangle with side =6
let the base be ABC and its centr be S.let the vertex of pyramid be V
draw od perpendicular to BC.join OB.In right angled triangle OBD
OB for equilateral triangle bisects BC and angle B.hence
BD=6/2 = 3
cos(B/2)=cos(60/2)=cos(30)=BD/OB
sqrt(3)/2 = 3/OB
OB = 3*2/sqrt(3) = 2sqrt(3)
now if we join OV,then OVB is a right angled triangle with
OV=height of pyramid = 5.2
OB=2sqrt(3)
OB^2=OV^2+OB^2=5.2^2+[2sqrt(3)]^2=39.04
OB = 6.25
OD/BD = tan(30)= 1/sqrt(3)
OD = BD/sqrt(3)=3/sqrt(3)=sqrt(3)
join VD....the slant height
in right triangle OVD
VD^2 = OV^2 + OD^2= 5.2^2+[sqrt(3)]^2=30.04
VD=5.5 = slant height
lateral surface area = 3*0.5*side*slant height
LSA=1.5*6*5.5=49.5
volume of pyramid = (1/3)*base area *ht
base area = [side^2][sqrt(3)/4]= 36*sqrt(3)/4=9sqrt(3)
volume = (1/3)9sqrt(3)*5.2=15.6sqrt(3)=27.02
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