# A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 2  and 8  and the pyramid's height is 2 . If one of the base's corners has an angle of (5pi)/6, what is the pyramid's surface area?

Oct 24, 2017

Total Surface Area  T S A = 34.8328

#### Explanation:

$C H = 2 \cdot \sin \left(5 \frac{\pi}{6}\right) = 1$
Area of parallelogram base $= 8 \cdot b 1 = 8 \cdot 1 = 8$

$E F = {h}_{1} = \sqrt{{h}^{2} + {\left(\frac{a}{2}\right)}^{2}} = \sqrt{{2}^{2} + {\left(\frac{8}{2}\right)}^{2}} = 4.4721$
Area of $\Delta A E D = B E C = \left(\frac{1}{2}\right) \cdot b \cdot {h}_{1} = \left(\frac{1}{2}\right) \cdot 2 \cdot 4.4721 = 4.4721$

$E G = {h}_{2} = \sqrt{{h}^{2} + {\left(\frac{b}{2}\right)}^{2}} = \sqrt{{2}^{2} + {\left(\frac{2}{2}\right)}^{2}} = 2.2361$
Area of $\Delta = C E D = A E C = \left(\frac{1}{2}\right) \cdot a \cdot {h}_{2} = \left(\frac{1}{2}\right) \cdot 8 \cdot 2.2361 = 8.9443$

Lateral surface area = $2 \cdot \Delta A E D + 2 \cdot \Delta C E D$
$= \left(2 \cdot 4.4721\right) + \left(2 \cdot 8.9443\right) = 26.8328$

Total surface area =Area of parallelogram base + Lateral surface area $= 8 + 26.8328 = 34.8328$

Oct 24, 2017

The pyramid's surface area is $8 + 10 \sqrt{21 + 4 \sqrt{3}}$

#### Explanation:

I recommend to use coordinates to solve this problem.

The pyramid's base is ABCD and its center is E

The coordinates of A and B are:
A(0;0)
B(8;0)

In order to find the coordinates of C and D, we need to use trigonometry and the pythagorean theorem:

$l = 2 \cos \left(\frac{\pi}{6}\right) = 2 \frac{\sqrt{3}}{2} = \sqrt{3}$
$L = \sqrt{{2}^{2} - {\sqrt{3}}^{2}} = \sqrt{4 - 3} = \sqrt{1} = 1$

Therefore the coordinates of C and D are:
C(8+sqrt(3);1)
D(sqrt(3);1)

A parallelogram's diagonals cross mid-length, thus the coordinates of E are:
E(4+sqrt(3)/2;1/2)

If the base is located in the (x;y) plane and the pyramid's peak (called F) is located at 2 above E, then the coordinates of all the pyramid's points are:
A(0;0;0)
B(8;0;0)
C(8+sqrt(3);1;0)
D(sqrt(3);1;0)
F(4+sqrt(3)/2;1/2;2)#

The pyramid's base's surface area is $A B \cdot L = 8 \cdot 1 = 8$

The ADF and BCF sides' surface areas are:
$\frac{A F \cdot A D}{2} = \frac{\sqrt{{\left(4 + \frac{\sqrt{3}}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2} + {2}^{2}} \cdot \cancel{2}}{\cancel{2}} = \sqrt{\left(16 + 4 \sqrt{3} + \frac{3}{4}\right) + \frac{1}{4} + 4} = \sqrt{21 + 4 \sqrt{3}} = A F$

The ABF and CDF sides' surface areas are:
$\frac{A F \cdot A B}{2} = A F \cdot 4 = 4 \sqrt{21 + 4 \sqrt{3}}$

Therefore, the pyramid's total surface area is:
$8 + 2 \sqrt{21 + 4 \sqrt{3}} + 8 \sqrt{21 + 4 \sqrt{3}} = 8 + 10 \sqrt{21 + 4 \sqrt{3}}$