AMSCO'S Geometry. New York - Rye High School

Pyramids 451For example, consider a regular pyramid withEsquare ABCD for a base and vertex E.The diagonals ofABCD intersect at M and AM BM CM DM.Since EM is perpendicular to the base, it is perpendicularto any line in the base through M. Therefore, D CEM ' MA and EMA is a right angle. Also,EM ' MB and EMB is a right angle. Since the diagonalsof a square are congruent and bisect each other,AMA > MB. Then since EM > EM, EMA EMBMBby SAS, and EA > EB because they are corresponding parts of congruent triangles.Similar reasoning will lead us to conclude that EB > EC, EC > ED, andED > EA. A similar proof can be given for any base that is a regular polygon.Therefore, we can make the following statement:E The lateral faces of a regular pyramid are isosceles triangles.In the regular pyramid with base ABCD and vertex E,ADBAB BC CD DA and AE BE CE DEC Therefore, ABE BCE CDE DAE, that is, the lateral faces ofthe pyramid are congruent. The lateral faces of a regular pyramid are congruent.EXAMPLE 2A regular pyramid has a base that is the hexagonABCDEF and vertex at V. If the length AB is 2.5 centimeters,and the slant height of the pyramid is 6 centimeters,find the lateral area of the pyramid.VSolutionThe slant height of the pyramid is the height of a lateralface. Therefore:Area of nABV 5 1 2 bhFE6 cmDC5 1 2 (2.5)(6)A 2.5 cm B5 15 2 cm2The lateral faces of the regular pyramid are congruent. Therefore, they haveequal areas. There are six lateral faces.Lateral area of the pyramid 5 6 A 152 B5 45 cm 2Answer