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?
Views of the solid
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^22*AB*AD*cos150^@#
#BD^2=4+492*2*7*(sqrt(3)/2)#
#BD^2=53+14sqrt3 => BD=sqrt(53+14sqrt3)#
#AC^2=AB^2+BC^22*AB*BC*cos30^@#
#AC^2=4+492*2*7*(sqrt3/2)#
#AC^2=5314sqrt3 => AC=sqrt(5314sqrt3)#
Finding the slant edge AE using
Finding the slant edge BD using
 Finding the height
#h_2# using#triangle_(ABE)# , 4th figure
#AE^2=h_2^2+m^2 => h_2^2=AE^2m^2#
#BE^2=h_2^2+(7m)^2#
#BE^2=AE^2cancel(m^2)+4914m+cancel(m^2)#
#14m=AE^2BE^2+49 => m=(7sqrt3)/2#
#> m^2=(4914sqrt3+3)/4=(5214sqrt3)/4#
#h_2^2=AE^2m^2=(249cancel(14sqrt3))/4(52cancel(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^2n^2#
#DE^2=h_3^2+(2+n)^2#
#DE^2=AE^2cancel(n^2)+4+4n+cancel(n^2)#
#4n=DE^2AE^24=(249+14sqrt3)/4(24914sqrt3)/44#
#n=7*sqrt3/41#
#> n^2=147/167*sqrt3/2+1 => n^2=163/167*sqrt3/2#
#h_3^2=AE^2n^2=(24914sqrt3)/4163/16+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#
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To find the surface area of the pyramid, you need to calculate the area of its base and the area of its lateral faces.

Area of the Base: Since the base is a parallelogram, you can find its area using the formula for the area of a parallelogram: ( \text{Area}\text{base} = \text{base} \times \text{height} ). Given that the base sides are 2 and 7, and the height is 7, the area of the base is ( \text{Area}\text{base} = 7 \times 7 = 49 ) square units.

Area of the Lateral Faces: The lateral faces of the pyramid are triangles. To find the area of each lateral face, you need to find the lengths of the sides of these triangles. Let's denote the length of the side of the base with length 2 as ( a ) and the length of the side of the base with length 7 as ( b ). The height of the pyramid is 7. The angle at the corner of the base is ( \frac{5\pi}{6} ), which is ( 150^\circ ). This means the angle at the apex of the pyramid is ( 30^\circ ). You can use trigonometry to find the lengths of the sides of the lateral faces.
For the side with length 2, the length of the adjacent side to the angle at the apex is ( 2 \cos(30^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} ). For the side with length 7, the length of the adjacent side to the angle at the apex is ( 7 \cos(30^\circ) = 7 \cdot \frac{\sqrt{3}}{2} = \frac{7\sqrt{3}}{2} ).
Now, using the formula for the area of a triangle ( A = \frac{1}{2} \times \text{base} \times \text{height} ), you can find the areas of the lateral faces.
 For the side with length 2: ( A_1 = \frac{1}{2} \times 2 \times 7 = 7 ) square units.
 For the side with length 7: ( A_2 = \frac{1}{2} \times 7 \times 7 = \frac{49}{2} ) square units.

Total Surface Area: The total surface area of the pyramid is the sum of the area of its base and the areas of its lateral faces: ( \text{Surface Area} = \text{Area}_\text{base} + 2 \times (A_1 + A_2) ). Plug in the values to find the total surface area.
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