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

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

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Answer 1 · 2016-06-10T23:44:57

$7 + \frac{7 \sqrt{197}}{2} + \frac{7 \sqrt{17}}{2} ≅ 70.555$ $7+(7sqrt(197))/2+(7sqrt(17))/2~=70.555$

Explanation:

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I created this figure using MS Excel

I created this figure using MS Excel

Data:
$A B = C D = 7$ $AB=CD=7$
$A D = B C = 2$ $AD=BC=2$
$E M = 7$ $EM=7$
$B ˆ A D = 150^{\circ}$ $B hat A D=150^@$

$S_{parallelogram} = b \cdot h_{1} = 7 \cdot 2 \cdot {cos 60}^{\circ} = 7 \cdot 1 = 7$ $S_("parallelogram")=b*h_1=7*2*cos60^@=7*1=7$

It can be proven that $A M = C M$ $AM=CM$ and $B M = D M$ $BM=DM$ (by using angle-side-angle). This is why:
$△_{A E M} \equiv △_{C E M} \Rightarrow A E = C E$ $triangle_(AEM) -= triangle_(CEM) => AE=CE$
and $△_{B E M} \equiv △_{D E M} \Rightarrow B E = D E$ $triangle_(BEM) -= triangle_(DEM)=> BE=DE$
what means that:
$△_{A B E} \equiv △_{C D E}$ $triangle_(ABE) -= triangle_(CDE)$
and $△_{A D E} \equiv △_{B C E}$ $triangle_(ADE) -= triangle_(BCE)$

So what we need to solve the problem is to find:
height of $△_{A B E}$ $triangle_(ABE)$ , $h_{2}$ $h_2$ in the 4th figure above
height of $△_{A D E}$ $triangle_(ADE)$ , $h_{3}$ $h_3$ in the 5th figure above

Finding the diagonals of the parallelogram (using Law of Cosines)
$B D^{2} = A B^{2} + A D^{2} - 2 \cdot A B \cdot A D \cdot {cos 150}^{\circ}$ $BD^2=AB^2+AD^2-2*AB*AD*cos150^@$
$B D^{2} = 4 + 49 - 2 \cdot 2 \cdot 7 \cdot (- \frac{\sqrt{3}}{2})$ $BD^2=4+49-2*2*7*(-sqrt(3)/2)$
$B D^{2} = 53 + 14 \sqrt{3} \Rightarrow B D = \sqrt{53 + 14 \sqrt{3}}$ $BD^2=53+14sqrt3 => BD=sqrt(53+14sqrt3)$

$A C^{2} = A B^{2} + B C^{2} - 2 \cdot A B \cdot B C \cdot {cos 30}^{\circ}$ $AC^2=AB^2+BC^2-2*AB*BC*cos30^@$
$A C^{2} = 4 + 49 - 2 \cdot 2 \cdot 7 \cdot (\frac{\sqrt{3}}{2})$ $AC^2=4+49-2*2*7*(sqrt3/2)$
$A C^{2} = 53 - 14 \sqrt{3} \Rightarrow A C = \sqrt{53 - 14 \sqrt{3}}$ $AC^2=53-14sqrt3 => AC=sqrt(53-14sqrt3)$

Finding the slant edge AE using $△_{A E M}$ $triangle_(AEM)$ in which $A M = \frac{A C}{2}$ $AM=(AC)/2$
$A E^{2} = E M^{2} + {(\frac{A C}{2})}^{2}$ $AE^2=EM^2+((AC)/2)^2$
$A E^{2} = 7^{2} + \frac{53 - 14 \sqrt{3}}{4} = \frac{196 + 53 - 14 \sqrt{3}}{4} = \frac{249 - 14 \sqrt{3}}{4}$ $AE^2=7^2+(53-14sqrt3)/4=(196+53-14sqrt3)/4=(249-14sqrt3)/4$

Finding the slant edge BD using $△_{B E M}$ $triangle_(BEM)$ in which $B M = \frac{B D}{2}$ $BM=(BD)/2$
$B E^{2} = E M^{2} + {(\frac{B D}{2})}^{2}$ $BE^2=EM^2+((BD)/2)^2$
$B E^{2} = 7^{2} + \frac{53 + 14 \sqrt{3}}{4} = \frac{196 + 53 + 14 \sqrt{3}}{4} = \frac{249 + 14 \sqrt{3}}{4}$ $BE^2=7^2+(53+14sqrt3)/4=(196+53+14sqrt3)/4=(249+14sqrt3)/4$

Finding the height $h_{2}$ $h_2$ using $△_{A B E}$ $triangle_(ABE)$ , 4th figure
$A E^{2} = h_{2}^{2} + m^{2} \Rightarrow h_{2}^{2} = A E^{2} - m^{2}$ $AE^2=h_2^2+m^2 => h_2^2=AE^2-m^2$
$B E^{2} = h_{2}^{2} + {(7 - m)}^{2}$ $BE^2=h_2^2+(7-m)^2$
$BE^2=AE^2-cancel(m^2)+49-14m+cancel(m^2)$
$14m=AE^2-BE^2+49 => m=(7-sqrt3)/2$
$-> m^2=(49-14sqrt3+3)/4=(52-14sqrt3)/4$

$h_2^2=AE^2-m^2=(249-cancel(14sqrt3))/4-(52-cancel(14sqrt3))/4$
$h_2^2=197/4 => h_2=sqrt197/2$
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:

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

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

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Explanation:

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