Question 34021
The base of a right pyramid is a regular hexagon with sides of length 12 m. The altitude of the pyramid is 9 m. Find the total surface area of the pyramid.					
Thanks for your help					
TOTAL SURFACE AREA=AREA OF HEXAGONAL BASE +6*AREA OF LATERAL TRIANGULAR FACE					
AREA OF HEXAGON=6*A^2*SQRT(3)/4=6*12^2*SQRT(3)/4=374.132					
LATERAL TRIANGULAR FACE HAS ALTITUDE OF 9 M. AND DISTANCE OF CENTRE FROM VERTEX OF BASE =12 M					
HENCE LATERAL EDGE = SQRT (12^2+9^2)=SQRT(144+81)=15					
AREA OF LATERAL TRIANGLE WITH SIDES 12,15 AND 15 IS GIVEN BY FORMULA SQRT{S(S-A)(S-B)(S-C)} WHERE S =(12+15+15)/2=21					
AREA=SQRT{21(21-12)(21-15)(21-15)}=SQRT(21*9*6*6)=18SQRT(21)=					
T.S.A.OF PYRAMID = 374.132+6*82.4864=869.0504