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A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is #3 #, its base has sides of length #2 #, and its base has a corner with an angle of # pi/4 #. What is the pyramid's surface area?

Answer 1

T S A = 19.0133

AB = BC = CD = DA = a = 2
Height OE = h = 3
OF = a/2 = 1/2 = 1
# EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(3^2+1^2) = color(red)(3.1623)#

Area of #DCE = (1/2)*a*EF = (1/2)*2*3.1623 = color(red)(3.1623)#
Lateral surface area #= 4*Delta DCE = 4*3.1623 = color(blue)(12.6492#)#

#/_C = pi/4, /_C/2 = pi/8#
diagonal #AC = d_1# & diagonal #BD = d_2#
#OB = d_2/2 = BCsin (C/2)=3sin(pi/8)= 1.1481

#OC = d_1/2 = BC cos (C/2) = 3* cos (pi/8) = 2.7716

Area of base ABCD #= (1/2)*d_1*d_2 = (1/2)(2*1.1481) (2*2.7716) = color (blue)(6.3641)#

T S A #= Lateral surface area + Base area#
T S A # =12.6492 + 6.3641 = color(purple)(19.0133)#

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Answer 2

James Atkins

The surface area of the pyramid is ( 12\sqrt{2} + 6\sqrt{3} ) square units.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.