A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $3$ $3$ and $7$ $7$ and the pyramid's height is $8$ $8$ . If one of the base's corners has an angle of $\frac{3 π}{8}$ $(3pi)/8$ , 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 $3$ $3$ and $7$ $7$ and the pyramid's height is $8$ $8$ . If one of the base's corners has an angle of $\frac{3 π}{8}$ $(3pi)/8$ , what is the pyramid's surface area?

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Answer 1 · 2018-06-13T20:54:26

$\frac{3}{2} \sqrt{305} + \frac{7}{4} \sqrt{1042 + 9 \sqrt{2}} = 83.0$ $3/2sqrt(305)+7/4sqrt(1042+9sqrt(2))=83.0$ to three significant figures

Explanation:

Work in the usual 3-dimensional Cartesian co-ordinate system $(x, y, z)$ $(x,y,z)$ ; let $(x, y, z) = (0, 0, 0)$ $(x,y,z)=(0,0,0)$ at the centre of the parallelogram base and the $x$ $x$ -axis run parallel to the long edge. Assume wlog that the smaller angle in the parallelogram is at positive $x$ $x$ and positive $y$ $y$ (or conversely negative both).

Then the apex of the pyramid is at $(0, 0, 8)$ $(0,0,8)$ and the mid-points of the parallelogram's four sides are at $(0, \pm \frac{7}{2}, 0)$ $(0,+-7/2,0)$ (short sides) and $(\pm \frac{3}{2} sin (\frac{3 π}{8}), 0, 0)$ $(+-3/2sin((3pi)/8),0,0)$ (long sides).

We calculate the area of the triangles on short and long sides by the usual triangle area formula: $\frac{1}{2} b h$ $1/2bh$ . $b$ $b$ is given in the question; $h$ $h$ is the slant height, the straight-line distance between side mid-point and apex.

Short sides:

$h^{2} = {(\frac{7}{2})}^{2} + 8^{2} = \frac{49}{4} + \frac{256}{4} = \frac{305}{4}$ $h^2=(7/2)^2+8^2=49/4+256/4=305/4$
$h = \sqrt{\frac{305}{4}} = \frac{\sqrt{305}}{2}$ $h=sqrt(305/4)=sqrt(305)/2$
$A_{s h} = \frac{1}{2} b h = \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{305}}{2} = \frac{3}{4} \sqrt{305}$ $A_{sh}=1/2bh=1/2*3*sqrt(305)/2=3/4sqrt(305)$

Long sides:

$h^{2} = {(\frac{3}{2})}^{2} {sin}^{2} (\frac{3 π}{8}) + 8^{2}$ $h^2=(3/2)^2sin^2((3pi)/8)+8^2$
We may take that $sin (\frac{3 π}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $sin((3pi)/8)=sqrt(2+sqrt(2))/2$ (proof here: https://www.mathway.com/popular-problems/Precalculus/400495), and so
$h^{2} = \frac{9}{4} \cdot \frac{2 + \sqrt{2}}{4} + 8^{2} = \frac{9}{16} (2 + \sqrt{2}) + 64 = \frac{1042 + 9 \sqrt{2}}{16}$ $h^2=9/4*(2+sqrt(2))/4+8^2=9/16(2+sqrt(2))+64=(1042+9sqrt(2))/16$
$h = \frac{\sqrt{1042 + 9 \sqrt{2}}}{4}$ $h=sqrt(1042+9sqrt(2))/4$
$A_{l o} = \frac{1}{2} b h = \frac{1}{2} \cdot 7 \cdot \frac{\sqrt{1042 + 9 \sqrt{2}}}{4} = \frac{7}{8} \sqrt{1042 + 9 \sqrt{2}}$ $A_{lo}=1/2bh=1/2*7*sqrt(1042+9sqrt(2))/4=7/8sqrt(1042+9sqrt(2))$

Thus the total surface area of the pyramid:

$A = 2 A_{s h} + 2 A_{l o} = \frac{3}{2} \sqrt{305} + \frac{7}{4} \sqrt{1042 + 9 \sqrt{2}}$ $A=2A_{sh}+2A_{lo}=3/2sqrt(305)+7/4sqrt(1042+9sqrt(2))$

To three significant figures, this equals 83.0 square units. Note that I have derived an exact expression here, but it is possible that the question setter desires everything to be computed on a calculator instead, in which case we do not need to know the $sin$ $sin$ formula used.

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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 $3$ $3$ and $7$ $7$ and the pyramid's height is $8$ $8$ . If one of the base's corners has an angle of $\frac{3 π}{8}$ $(3pi)/8$ , 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 $3$ $3$ and $7$ $7$ and the pyramid's height is $8$ $8$ . If one of the base's corners has an angle of $\frac{3 π}{8}$ $(3pi)/8$ , what is the pyramid's surface area?