Question

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 `9`, its base's sides have lengths of `5`, and its base has a corner with an angle of `(5 pi)/8`. What is the pyramid's surface area?

Answer 1

T S A = 111.0851

Explanation:

AB = BC = CD = DA = a = 5
Height OE = h = 9
OF = a/2 = 5/2 = 2.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(9^2+(2.5)^2) = color(red)9.3408`

Area of `DCE = (1/2)*a*EF = (1/2)*5*9.3408 = color(red)(23.3519)`
Lateral surface area `= 4*Delta DCE = 4*23.3519 = color(blue)(93.4076)`

`/_C =5 pi/8, /_C/2 = 3pi/8`
diagonal `AC = d_1` & diagonal `BD = d_2`
`OB = d_2/2 = BC*sin (C/2)=5*sin(3pi/8) = **4.6194**`

#OC = d_1/2 = BC cos (C/2) = 5cos (3pi/8) = 1.9134*

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*4.6194)(2*1.9134) = color (blue)(17.6775)`

Total Surface Area `= Lateral surface area + Base area`
T S A `=93.4076 + 17.6775 = color(purple)(111.0851)`