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
Explanation:
Views of the solid
I created this figure using MS Excel
I created this figure using MS Excel
I created this figure using MS Excel
I created this figure using MS Excel
I created this figure using MS Excel
Data:
It can be proven that
and
what means that:
and
So what we need to solve the problem is to find:
height of
height of
- 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
Finding the slant edge BD using
- Finding the height
h_2 usingtriangle_(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 findh_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 usingtriangle_(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 findh_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