document.write( "Question 62559: for homework there is a challenge question that says:\r
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document.write( "This net will fold into a triangular pyramid. Find the surface area of the pyramid, and estimate its volume.
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document.write( "The height of the pyramid is 5.2cm and the length is 6cm \r
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document.write( "I got this from the Math Impact Textbook, course 2(purple book) \n" );
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Algebra.Com's Answer #43359 by venugopalramana(3286)![]() ![]() 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( " |