A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 2 and 7 and the pyramid's height is 7 . If one of the base's corners has an angle of (5pi)/6, what is the pyramid's surface area?

1 Answer
Jun 10, 2016

7+(7sqrt(197))/2+(7sqrt(17))/2~=70.555

Explanation:

Views of the solid
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Data:
AB=CD=7
AD=BC=2
EM=7
B hat A D=150^@

S_("parallelogram")=b*h_1=7*2*cos60^@=7*1=7

It can be proven that AM=CM and BM=DM (by using angle-side-angle). This is why:
triangle_(AEM) -= triangle_(CEM) => AE=CE
and triangle_(BEM) -= triangle_(DEM)=> BE=DE
what means that:
triangle_(ABE) -= triangle_(CDE)
and triangle_(ADE) -= triangle_(BCE)

So what we need to solve the problem is to find:
height of triangle_(ABE), h_2 in the 4th figure above
height of triangle_(ADE), h_3 in the 5th figure above

  • Finding the diagonals of the parallelogram (using Law of Cosines)
    BD^2=AB^2+AD^2-2*AB*AD*cos150^@
    BD^2=4+49-2*2*7*(-sqrt(3)/2)
    BD^2=53+14sqrt3 => BD=sqrt(53+14sqrt3)

AC^2=AB^2+BC^2-2*AB*BC*cos30^@
AC^2=4+49-2*2*7*(sqrt3/2)
AC^2=53-14sqrt3 => AC=sqrt(53-14sqrt3)

Finding the slant edge AE using triangle_(AEM) in which AM=(AC)/2
AE^2=EM^2+((AC)/2)^2
AE^2=7^2+(53-14sqrt3)/4=(196+53-14sqrt3)/4=(249-14sqrt3)/4

Finding the slant edge BD using triangle_(BEM) in which BM=(BD)/2
BE^2=EM^2+((BD)/2)^2
BE^2=7^2+(53+14sqrt3)/4=(196+53+14sqrt3)/4=(249+14sqrt3)/4

  • Finding the height h_2 using triangle_(ABE), 4th figure
    AE^2=h_2^2+m^2 => h_2^2=AE^2-m^2
    BE^2=h_2^2+(7-m)^2
    BE^2=AE^2-cancel(m^2)+49-14m+cancel(m^2)
    14m=AE^2-BE^2+49 => m=(7-sqrt3)/2
    -> m^2=(49-14sqrt3+3)/4=(52-14sqrt3)/4

h_2^2=AE^2-m^2=(249-cancel(14sqrt3))/4-(52-cancel(14sqrt3))/4
h_2^2=197/4 => h_2=sqrt197/2
We can also find h_2 in this way:
h_2^2=(h_1/2)^2+EM^2=(1/2)^2+7^2=1/4+49=197/4 => h_2=sqrt197/2

  • Finding the height h_3 using triangle_(ADE), 5th figure
    AE^2=h_3^2+n^2 => h_3^2=AE^2-n^2
    DE^2=h_3^2+(2+n)^2
    DE^2=AE^2-cancel(n^2)+4+4n+cancel(n^2)
    4n=DE^2-AE^2-4=(249+14sqrt3)/4-(249-14sqrt3)/4-4
    n=7*sqrt3/4-1
    -> n^2=147/16-7*sqrt3/2+1 => n^2=163/16-7*sqrt3/2

h_3^2=AE^2-n^2=(249-14sqrt3)/4-163/16+7*sqrt3/2
h_3^2=833/16 => h_3=(7sqrt17)/4
We can also find h_3 in this way:
h_3^2=( (7cos60^@)/2)^2+EM^2=(7/4)^2+7^2=49/16+49=833/16 => h_3=(7sqrt17)/4

Finally:

S_T=S_"paralellogram" +2*S_(triangle_(ABE))+2*S_(triangle_(ADE))
S_T=7+cancel(2)*(7*sqrt197/2)/cancel(2)+cancel(2)*(cancel(2)*7*sqrt17/cancel(4))/2
S_T=7+(7sqrt197)/2+(7sqrt17)/2