document.write( "Question 62559: for homework there is a challenge question that says:\r
\n" ); document.write( "\n" ); document.write( "This net will fold into a triangular pyramid. Find the surface area of the pyramid, and estimate its volume.
\n" ); document.write( "
\n" ); document.write( "The height of the pyramid is 5.2cm and the length is 6cm \r
\n" ); document.write( "\n" ); document.write( "I got this from the Math Impact Textbook, course 2(purple book)
\n" ); document.write( "

Algebra.Com's Answer #43359 by venugopalramana(3286)\"\" \"About 
You can put this solution on YOUR website!
for homework there is a challenge question that says:
\n" ); document.write( "This net will fold into a triangular pyramid. Find the surface area of the pyramid, and estimate its volume.\r
\n" ); document.write( "\n" ); document.write( "The height of the pyramid is 5.2cm and the length is 6cm
\n" ); document.write( "I got this from the Math Impact Textbook, course 2(purple book)
\n" ); document.write( "base of triangular pyramid given is an equilateral triangle with side =6
\n" ); document.write( "let the base be ABC and its centr be S.let the vertex of pyramid be V
\n" ); document.write( "draw od perpendicular to BC.join OB.In right angled triangle OBD
\n" ); document.write( "OB for equilateral triangle bisects BC and angle B.hence
\n" ); document.write( "BD=6/2 = 3
\n" ); document.write( "cos(B/2)=cos(60/2)=cos(30)=BD/OB
\n" ); document.write( "sqrt(3)/2 = 3/OB
\n" ); document.write( "OB = 3*2/sqrt(3) = 2sqrt(3)
\n" ); document.write( "now if we join OV,then OVB is a right angled triangle with
\n" ); document.write( "OV=height of pyramid = 5.2
\n" ); document.write( " OB=2sqrt(3)
\n" ); document.write( "OB^2=OV^2+OB^2=5.2^2+[2sqrt(3)]^2=39.04
\n" ); document.write( "OB = 6.25
\n" ); document.write( "OD/BD = tan(30)= 1/sqrt(3)
\n" ); document.write( "OD = BD/sqrt(3)=3/sqrt(3)=sqrt(3)
\n" ); document.write( "join VD....the slant height
\n" ); document.write( "in right triangle OVD
\n" ); document.write( "VD^2 = OV^2 + OD^2= 5.2^2+[sqrt(3)]^2=30.04
\n" ); document.write( "VD=5.5 = slant height
\n" ); document.write( "lateral surface area = 3*0.5*side*slant height
\n" ); document.write( "LSA=1.5*6*5.5=49.5 \r
\n" ); document.write( "\n" ); document.write( "volume of pyramid = (1/3)*base area *ht
\n" ); document.write( "base area = [side^2][sqrt(3)/4]= 36*sqrt(3)/4=9sqrt(3)
\n" ); document.write( "volume = (1/3)9sqrt(3)*5.2=15.6sqrt(3)=27.02
\n" ); document.write( "
\n" );