Question

Can someone find the volume of this pyramid?

Answer 1

Check the picture below.

now, let's notice the triangle on the base.... is an isosceles, two sides are twins, and therefore, the two angles they make out are also twins, so, if the central angle is 100°, the other angles are each 80°/2, namely 40°, therefore, using the law of sines,

`\bf \textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{sin(100^o)}{2.5}=\cfrac{sin(40^o)}{r}\implies r\cdot sin(100^o)=2.5\cdot sin(40^o) \\\\\\ r=\cfrac{2.5\cdot sin(40^o)}{sin(100^o)}\implies \boxed{r\approx 1.63}`

now, let's notice the shaded triangle, namely the one with 35°, is a right-triangle, namely one angle is 35°, another 90°, so the last one must be 55°.

the opposite side to the 35° is "r", which we know is about 1.63, so again, let's use the law of sines to find the side "h",

`\bf \cfrac{r}{sin(35^o)}=\cfrac{h}{sin(55^o)}\implies r\cdot sin(55^o)=h\cdot sin(35^o) \\\\\\ \cfrac{r\cdot sin(55^o)}{sin(35^o)}=h\implies \cfrac{1.63\cdot sin(55^o)}{sin(35^o)}\approx h\implies \boxed{2.33\approx h} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a circular cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r\approx 1.63\\ h\approx 2.33 \end{cases}\implies V\approx \cfrac{\pi (1.63)^2(2.33)}{3}\implies \blacktriangleright V\approx 6.48 \blacktriangleleft`

make sure your calculator is in Degree mode.