# A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is 8 , its base has sides of length 6 , and its base has a corner with an angle of (3 pi)/4 . What is the pyramid's surface area?

##### 1 Answer
Oct 11, 2017

Surface area of pyramid with rhombus base = 127.9839

#### Explanation:

Opposite base angles of the rhombus are (3pi)/4 & pi/4 respy.
d1 & d2 be the two diagonals intersecting at right angle.
$\sin \left(\frac{\theta}{2}\right) = \sin \left(\frac{\frac{\pi}{4}}{2}\right) = \sin \left(\frac{\pi}{8}\right) = \left(\frac{d \frac{1}{2}}{6}\right)$
$\sin \left(\frac{\pi}{8}\right) = d \frac{1}{12} \mathmr{and} d 1 = 12 \cdot \sin \left(\frac{\pi}{8}\right) =$4.5922

Similarly,$\cos \left(\frac{\pi}{8}\right) = \left(\frac{d \frac{2}{6}}{2}\right) = \left(d \frac{2}{12}\right)$
$d 2 = 12 \cdot \cos \left(\frac{\pi}{8}\right) =$11.0866

Area of Rhombus base $= \frac{d 1 \cdot d 2}{2} = \frac{4.5922 \cdot 11.0866}{2} =$25.4559

Height of one side surface $h 1 = \sqrt{{\left(\frac{b}{2}\right)}^{2} + {h}^{2}}$
As $\frac{b}{2} = \frac{6}{2} = 3 , h 1 = \sqrt{{3}^{2} + {8}^{2}} = 8.544$

Area of 4 sides of pyramid $= 4 \cdot \left(\frac{1}{2}\right) b \cdot h 1 = 2 \cdot 6 \cdot 8.544 =$102.588

Surface area of pyramid = base area + 4 side areas
#=25.4559+102.588=127.9839 